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Econ-163 · economics · computer science · Algorithmic game theory · Computational economics

Explicit constant-regret bounds for plain Optimistic Hedge

LLM status Partially solved unverified

Statement

Explicit constant-regret bounds for plain Optimistic Hedge

Status checked: 17 October 2026

Status

This is a literature-motivated quantitative problem. Existence of some dimension-dependent constants is already claimed by a preprint; this proposal concerns effective bounds for the specified update.

Game, feedback, and regret

Fix integers $N\geq 2$ and $K\geq 2$. A game has labelled players $i\in\{1,\ldots,N\}$, action sets $A_i=\{1,\ldots,m_i\}$ with $1\leq m_i\leq K$, and payoffs $u_i:\prod_j A_j\to[0,1]$. Payoffs extend multilinearly to $\Sigma=\prod_i\Delta(A_i)$. At round $t\geq1$, all players choose mixed strategies simultaneously. After play, player $i$ receives its own exact vector $v_i^t(a)=u_i(a,x_{-i}^t)$ for every $a\in A_i$. An uncoupled update uses its own past vectors, its state, its own label and the public dimensions, without access to any player's payoff table or to precomputed equilibria. All logarithms are natural. Set \[ R_i(T)=\max_{a\in A_i}\sum_{t=1}^T \left(v_i^t(a)-\langle x_i^t,v_i^t\rangle\right), \qquad [r]_+=\max\{r,0\}. \] The comparator is one fixed action. A horizon-uniform bound concerns $\sup_T R_i(T)$, rather than $\sup_T|R_i(T)|$ or a comparator that changes actions each round. This is expected-payoff feedback and mixed-payoff regret, rather than realized regret after action sampling.

The fixed algorithm

Initialize $x_{ia}^1=1/m_i$ and $v_i^0=0$. For a common constant step size $\eta>0$, use exactly \[ x_{ia}^{t+1}=\frac{x_{ia}^{t}\exp(\eta(2v_{ia}^{t}-v_{ia}^{t-1}))} {\sum_{b\in A_i}x_{ib}^{t}\exp(\eta(2v_{ib}^{t}-v_{ib}^{t-1}))}. \] The payoff form agrees with the loss form under $\ell_i=1-v_i$; action-independent factors cancel in normalization, including the first step with zero initial feedback.

Question R2

Find explicit computable functions $\eta(N,K)>0$ and $B(N,K)<\infty$ such that for every allowed action-count vector, every game, every player, and every horizon, the simultaneous iterates satisfy \[ R_i(T)\leq B(N,K). \] A precise quantitative subquestion is whether there are absolute constants $c,p,q>0$ and an explicit such step-size rule with \[ B(N,K)\leq cN^p(1+\log K)^q \quad(N,K\geq2). \] A negative answer to that polynomial question must concern every step-size rule for the displayed algorithm, rather than one unsuitable choice of $\eta$. A useful first step is any fully effective finite bound. Matching the $O(\sqrt N\log K)$ scale known for a different algorithm is a stronger comparison, not a premise.

Known claim and unresolved quantitative issue

Ha reports that for each action-count vector there exist $\eta_\star>0$ and $C_\star<\infty$, independent of the payoff table and horizon, such that for $0<\eta\leq\eta_\star$ \[ R_i(T)\leq\frac{\log m_i}{\eta}+C_\star\eta. \] The text after Corollary 4.2 and the conclusion explicitly leave dimension-explicit constants for future work [ha, Theorem 4.1]. The proposed $B$ would have to be uniform over all action counts at most $K$; there are finitely many such vectors for fixed $N,K$. The source statement was checked; its full proof has not been independently audited here.

Duplicate and scope checks

No exact we-solve target for effective plain-Optimistic-Hedge constants was found. ECON-C73-63 permits a larger learner class but imposes additional adversarial and computation restrictions; this proposal is not claimed to settle it. The algorithm above is plain Optimistic Hedge, not a modified higher-order or regret-matching rule. The model does not include sampled utilities, floating-point rounding, or last-iterate Nash convergence.

Preparation

The original research map was generated using OpenAI GPT 6 Astra Pro. Source checks, formulation, and LaTeX preparation used a Codex assistant based on GPT-6. Exact model versions and reasoning settings are unknown. The cited researchers retain credit for their results. Source statements and scope were checked; no complete proof audit or novelty certification is claimed.
Origin of the questions.
The open-problem suggestions motivating this entry are credited to Rida Laraki's Facebook comment [laraki-facebook].

References

  • [ha] Junsoo Ha. A Horizon-Independent Regret Bound for Optimistic Hedge in General-Sum Games. arXiv:2609.22839v1, 19 September 2026. Equation $5$; Theorem 4.1; text after Corollary 4.2; conclusion. https://arxiv.org/html/2609.22839v1.
  • [laraki-facebook] Rida Laraki. Facebook comment suggesting open problems in game theory.

Statement licence: CC-BY-4.0 · Source file

Literature source

ARXIV: 2609.22839v1

Theorem 4.1; discussion after Corollary 4.2; conclusion

Citation not independently checked.

Contributions 1

Econ-163-A1 · Partial result · LLM-generated · gpt 6 astra pro

Explicit special cases and a quantitative recurrence criterion for plain Optimistic Hedge

Explicit special cases and a quantitative recurrence criterion for plain Optimistic Hedge

OpenAI GPT 6 Astra Pro

18 October 2026

Problem

WS-107791: Explicit constant-regret bounds for plain Optimistic Hedge. The downloaded submission prompt identifies this entry as ECON-2; the permanent identifier above is used throughout this contribution. Public problem page.

Title

Explicit special cases and a quantitative recurrence criterion for plain Optimistic Hedge.

Claim type

Partial result, with a conditional general-game criterion. The full problem remains unresolved.

Scope

The exact algorithm in the problem admits: (i) the bound $R_i(T)\leq4\log K$ at $\eta=\log2$ when player $i$ has a weakly dominant pure action, even against arbitrary opponent play; (ii) the bound $R_i(T)\leq(4N-2)\log K+3/2$ at $\eta=1/[2(N-1)]$ in the class specified by (eq:aggregate-condition), which includes all two-player constant-sum games; and (iii) an entirely numerical conversion from supplied finite recurrence parameters to a horizon-uniform regret bound. A static-payoff example also proves a lower bound $(\log K)/\eta-3/2$ for any proposed step size. Section 6 conditionally computes general-game recurrence parameters under decidability of the real exponential field. No unconditional general-game parameter rule is supplied.

Assumptions and limitations

There are $N\geq2$ players, $1\leq m_i\leq K$, and payoffs in $[0,1]$. The feedback is the exact expected payoff vector and the regret is mixed-payoff external regret against one fixed action. All logarithms are natural. No sampled feedback, realized-action regret, floating-point implementation, or last-iterate convergence is asserted. The special cases are included as self-contained quantitative benchmarks, not claimed as new qualitative resolutions of their previously studied classes. The recurrence criterion is a quantitative development of Ha's method [ha]. Section 6 gives a terminating procedure to compute general-game bounds conditional on decidability of the real exponential field. That additional assumption is not proved here, and no running-time or polynomial bound is asserted.

Write-up

1. Exact update and the outstanding question

The update is precisely \begin{equation}\label{wsref-0-65-71-3a-75-70-64-61-74-65} x_{ia}^{t+1}=\frac{x_{ia}^t\exp\{\eta(2v_{ia}^t-v_{ia}^{t-1})\}} {\sum_bx_{ib}^t\exp\{\eta(2v_{ib}^t-v_{ib}^{t-1})\}}, \qquad x_{ia}^1=1/m_i,\quad v_i^0=0. \end{equation} Its telescoped representation is \begin{equation}\label{wsref-0-65-71-3a-74-65-6c-65-73-63-6f-70-65-64} x_{ia}^t\ \propto\ \exp\!\left\{\eta\left(\sum_{s<t}v_{ia}^s+v_{ia}^{t-1}\right)\right\}. \end{equation} The exponent at $t=1$ is zero. At $t\geq2$, each earlier vector through $t-2$ has coefficient one and the vector at $t-1$ has coefficient two. This checks the initialization and the one-step optimistic term.

Ha's Theorem 4.1 asserts that for each action-count vector, some positive step-size interval and some finite constant give $R_i(T)\leq(\log m_i)/\eta+C_\star\eta$ uniformly in the game and horizon. The text following Corollary 4.2 leaves dimension-explicit bounds open [ha]. Its finite-recurrence argument was inspected in Sections 5.3--5.5 and Appendices B--D. No contradiction to that argument was found in this check. This is not a claim to have reconstructed every historical analytic result on which its Noetherianity step relies.

2. Weakly dominant pure actions: a uniform explicit bound

Theorem: Suppose player $i$ has a pure action $a_\star$ such that $u_i(a_\star,a_{-i})\geq u_i(a,a_{-i})$ for every $a,a_{-i}$. For every $\eta>0$, every opponent strategy sequence, and every $T$, \[ R_i(T)\leq\frac{e^\eta\log m_i}{1-e^{-\eta}}. \] Consequently, if each player has such an action, the common explicit choice $\eta=\log2$ gives $R_i(T)\leq4\log K$ for every player and horizon. No strictly positive dominance gap is required.
Proof: Suppress $i$. Put $d_a^t=v^t(a_\star)-v^t(a)$ and $d_a^0=0$. Weak dominance and normalization imply $0\leq d_a^t\leq1$ and $d_{a_\star}^t=0$. The action $a_\star$ maximizes cumulative payoff at every horizon, so $R(T)=\sum_{t\leq T}\langle x^t,d^t\rangle$. Let $D_a^{t-1}=\sum_{s<t}d_a^s$, and define \[ W_t=\sum_a e^{-\eta D_a^{t-1}},\qquad y_a^t=\frac{e^{-\eta D_a^{t-1}}}{W_t}. \] Equation (eq:telescoped) gives $x_a^t=y_a^t e^{-\eta d_a^{t-1}}/\sum_b y_b^t e^{-\eta d_b^{t-1}}$. The denominator is at least $e^{-\eta}$ and the numerator factor at most one, hence $x_a^t\leq e^\eta y_a^t$. Therefore $\langle x^t,d^t\rangle\leq e^\eta\langle y^t,d^t\rangle$.

Convexity of $z\mapsto e^{-\eta z}$ on $[0,1]$ places its graph below its endpoint chord: $e^{-\eta z}\leq1-(1-e^{-\eta})z$. Thus \[ \frac{W_{t+1}}{W_t}=\sum_a y_a^t e^{-\eta d_a^t} \leq1-(1-e^{-\eta})\langle y^t,d^t\rangle. \] Since $-\log(1-z)\geq z$ for $0\leq z<1$, \[ (1-e^{-\eta})\sum_{t=1}^T\langle y^t,d^t\rangle \leq\log W_1-\log W_{T+1}\leq\log m. \] The last inequality uses $W_1=m$ and the unit contribution of $a_\star$ to $W_{T+1}$. Combining the two estimates proves the result, including $m=1$, where the regret is zero. ∎

3. A general aggregate-comparator condition

Suppose there is a fixed mixed-strategy profile $p=(p_1,\ldots,p_N)$ such that \begin{equation}\label{wsref-0-65-71-3a-61-67-67-72-65-67-61-74-65-2d-63-6f-6e-64-69-74-69-6f-6e} \sum_{i=1}^N\left[u_i(p_i,x_{-i})-u_i(x)\right]\geq0 \qquad\text{for every mixed profile }x. \end{equation} The learners need not know $p$. Its existence is used only in the proof. Condition (eq:aggregate-condition) is an additional game-class assumption; it is not a property of every general-sum game.
Theorem: Under (eq:aggregate-condition), write $L=\sum_j\log m_j$. If $0<\eta\leq1/[2(N-1)]$, the exact update (eq:update) satisfies \begin{equation}\label{wsref-0-65-71-3a-61-67-67-72-65-67-61-74-65-2d-62-6f-75-6e-64} R_i(T)\leq\frac{\log m_i}{\eta} +\eta\left[1+4(N-1)L+4\eta^2N(N-1)\right] \quad\text{for every }i,T. \end{equation} In particular, $\eta=1/[2(N-1)]$ yields $R_i(T)\leq(4N-2)\log K+3/2$.
Proof: For each player, define $y_i^t\propto\exp(\eta\sum_{s<t}v_i^s)$ and set \[ a_{it}=\lVert y_i^{t+1}-x_i^t\rVert_1,\qquad b_{it}=\lVert x_i^t-y_i^t\rVert_1,\qquad d_{it}=\lVert v_i^t-v_i^{t-1}\rVert_\infty. \] All coordinates of $x_i^t,y_i^t$ are positive at finite $t$. For distributions $z,w$, let \[ D(z\Vert w)=\sum_a z_a\log(z_a/w_a), \] with the usual $0\log0=0$ convention. Exponential weights give the following identity for every comparator $q_i$: \begin{align*} \eta\langle q_i-x_i^t,v_i^t\rangle ={}&D(q_i\Vert y_i^t)-D(q_i\Vert y_i^{t+1}) -D(y_i^{t+1}\Vert x_i^t)-D(x_i^t\Vert y_i^t)\\ &+\eta\langle y_i^{t+1}-x_i^t,v_i^t-v_i^{t-1}\rangle. \end{align*} To verify it, apply the log-ratio identities for $y_i^{t+1}\propto y_i^t e^{\eta v_i^t}$ and $x_i^t\propto y_i^t e^{\eta v_i^{t-1}}$, and add them. Pinsker's inequality $D(z\Vert w)\geq\lVert z-w\rVert_1^2/2$ and $\eta a d\leq a^2/4+\eta^2d^2$ imply, after summing in time and maximizing over $q_i$, \begin{equation}\label{wsref-0-65-71-3a-72-76-75-2d-6c-31} R_i(T)\leq\frac{\log m_i}{\eta} +\eta\sum_{t\leq T}d_{it}^2 -\frac1{4\eta}\sum_{t\leq T}(a_{it}^2+b_{it}^2). \end{equation} This is the standard optimistic mirror-descent calculation, with constants given here explicitly; predictable-sequence methods are developed in [rs].

Put $E_T=\sum_{i,t\leq T}(a_{it}^2+b_{it}^2)$ and $\delta_{jt}=\lVert x_j^t-x_j^{t-1}\rVert_1$ for $t\geq2$. Bounded payoffs and total variation of product distributions give \[ d_{it}\leq\frac12\sum_{j\ne i}\delta_{jt}\quad(t\geq2), \qquad d_{i1}\leq1. \] Indeed, changing one opponent's distribution changes the expectation of a $[0,1]$-valued payoff by at most half its $\ell_1$ change, and replacing opponents one at a time proves the displayed estimate. Also $\delta_{jt}\leq b_{jt}+a_{j,t-1}$, so $\sum_{j,t=2}^T\delta_{jt}^2\leq2E_T$. Consequently, \begin{equation}\label{wsref-0-65-71-3a-61-6c-6c-2d-66-65-65-64-62-61-63-6b} \sum_{i,t\leq T}d_{it}^2 \leq N+\frac{(N-1)^2}{2}E_T, \qquad \sum_{t\leq T}d_{it}^2 \leq1+\frac{N-1}{2}E_T. \end{equation}

The external-regret maximum is at least the regret against each $p_i$. Summing (eq:aggregate-condition) over rounds therefore proves $\sum_iR_i(T)\geq0$. Summing (eq:rvu-l1) and applying (eq:all-feedback), \[ 0\leq\sum_iR_i(T) \leq\frac L\eta+\eta N+ \left(\frac{\eta(N-1)^2}{2}-\frac1{4\eta}\right)E_T. \] The step-size restriction makes the coefficient at most $-1/(8\eta)$. Hence $E_T\leq8L+8\eta^2N$. Substitute this into the individual bound in (eq:all-feedback), then drop the negative term in (eq:rvu-l1), to obtain (eq:aggregate-bound). At the stated explicit step size, its right side is at most \[ (4N-2)\log K+\frac1{2(N-1)}+\frac{N}{2(N-1)^2} \leq(4N-2)\log K+\frac32. \] ∎

Corollary: In every two-player constant-sum game with each payoff in $[0,1]$, plain Optimistic Hedge with $\eta=1/2$ satisfies $R_i(T)\leq6\log K+3/2$ for all $i,T$.
Proof: Let $u_1+u_2=c$, and let $p_1,p_2$ be minimax strategies for $u_1$, with value $V$. For every $x_1,x_2$, $u_1(p_1,x_2)\geq V\geq u_1(x_1,p_2)$. Therefore \[ \sum_{i=1}^2[u_i(p_i,x_{-i})-u_i(x)] =u_1(p_1,x_2)-u_1(x_1,p_2)\geq0. \] Apply Theorem Theorem with $N=2$. ∎
The qualitative constant-regret phenomenon in this class was already known. Tsuchiya studies substantially sharper action-count dependence and constants [tsuchiya]; the corollary is a directly checked benchmark for the exact initialization in this problem.

4. A lower bound that every proposed step-size rule must respect

Proposition: For every $N,K\geq2$ and every fixed $\eta>0$, there is an allowed game in which \[ \sup_{T\geq1}R_i(T)\geq\frac{\log K}{\eta}-\frac32. \] Thus any uniform bound for a proposed rule $\eta(N,K)$ must be at least this quantity (as well as nonnegative).
Proof: Give player $i$ exactly $K$ actions, with payoff one for one action and zero for all the others, independently of the opponents. Give all other players arbitrary allowed action sets and constant payoffs. Put $q=K-1$. At $t=1$ the regret increment is $q/K$. At $t\geq2$, (eq:telescoped) gives the good action probability $e^{\eta t}/(e^{\eta t}+q)$, so \[ \sup_T R_i(T)=\frac qK+\sum_{t=2}^\infty\frac q{e^{\eta t}+q}. \] The summand is decreasing, and comparison with its integral gives \[ \sum_{t=2}^\infty\frac q{e^{\eta t}+q} \geq\int_2^\infty\frac q{e^{\eta s}+q}\,ds =\frac1\eta\log(1+q e^{-2\eta}) \geq\frac{\log K}{\eta}-2. \] Since $q/K\geq1/2$, the claimed bound follows. ∎
This example rules out making an extremely small step size costless. It does not disprove a polynomial bound: that would require controlling all possible dimension-dependent step-size rules.

5. Making the finite-recurrence argument quantitative

The next theorem replaces all constants after the recurrence step by explicit formulas. It identifies a finite set of quantitative data that would suffice to settle the general problem.

Use losses $\ell_i^t=1-v_i^t$ for $t\geq1$, with $\ell_i^0=0$. This produces the same normalized update as (eq:update); the action-independent first-step factor cancels. For $a<b$ define \[ h_{i,ab}^t=\sqrt{x_{ia}^tx_{ib}^t} (\ell_{ia}^{t-1}-\ell_{ib}^{t-1}),\qquad \Delta z^t=z^{t+1}-z^t. \] In particular $h_i^1=0$, and the pairwise variance identity gives $\lVert h_i^t\rVert_2^2=\operatorname{Var}_{x_i^t}(\ell_i^{t-1})\leq1/4$ for $t\geq2$.

Theorem (Quantitative recurrence criterion): Suppose numerical values $m\geq1$, $A,M\geq0$, and $\eta_0>0$ have been proved to work for every player and every game in a specified class, in the following sense: for all $0<\eta\leq\eta_0$ and all $t\geq1$, \begin{align} \Delta^m h_i^{t+m} &=\sum_{k=0}^{m-1}c_{ikt}\eta^{m-k}\Delta^k h_i^{t+k}, & |c_{ikt}|&\leq A,\label{wsref-0-65-71-3a-63-65-72-74-69-66-69-63-61-74-65-2d-72-65-63-75-72-72-65-6e-63-65}\\ \lVert \Delta^k h_i^{t+k}\rVert_2&\leq M\eta^k, &0\leq k&\leq m.\label{wsref-0-65-71-3a-63-65-72-74-69-66-69-63-61-74-65-2d-73-6d-6f-6f-74-68} \end{align} The coefficients may depend on the player, game, time, and step size. Define explicit quantities \[ P=2m\max\{A,1\},\quad E=3^mM\sqrt{8m},\quad D=16mM^2,\quad Q=2(2E)^{2/m},\quad C=2P^2+Q, \] and \[ \bar\eta=\min\!\left\{\eta_0,\frac1{12}, \frac1{\sqrt{6(C+32)}}\right\}. \] For every $0<\eta\leq\bar\eta$, every player, and every horizon, \begin{equation}\label{wsref-0-65-71-3a-63-65-72-74-69-66-69-63-61-74-65-2d-62-6f-75-6e-64} R_i(T)\leq\frac{\log m_i}{\eta} +\eta\left(D+Q+\frac1{24}\right). \end{equation} Thus supplied uniform recurrence data produce explicit functions $\eta=\bar\eta$ and $B=(\log K)/\bar\eta+\bar\eta(D+Q+1/24)$ for that class.

5.1. An explicit cutoff lemma

Lemma: Let $0<\eta\leq1$ and let a Euclidean-valued sequence $(u^t)_{t\geq1}$ satisfy (eq:certificate-recurrence)--(eq:certificate-smooth) with $h_i$ replaced by $u$. For every $T\geq1$, there is a finitely supported sequence $y$ on $\mathbb Z$ such that, writing $U_j=\lVert \Delta^j y\rVert_{\ell_2}$, \begin{align} U_0^2&\leq\sum_{t=1}^{T+1}\lVert u^t\rVert_2^2,\label{wsref-0-65-71-3a-63-75-74-6f-66-66-2d-7a-65-72-6f}\\ \sum_{t=1}^T\lVert \Delta u^t\rVert_2^2&\leq U_1^2+D,\label{wsref-0-65-71-3a-63-75-74-6f-66-66-2d-6f-6e-65}\\ U_m&\leq A\sum_{k=0}^{m-1}\eta^{m-k}U_k +E\eta^{m-1/2}.\label{wsref-0-65-71-3a-63-75-74-6f-66-66-2d-68-69-67-68} \end{align} Here $D=16mM^2$ and $E=3^mM\sqrt{8m}$.
Proof: Set $J=\lceil1/\eta\rceil$ and $w=m(J-1)$. If $T\leq2w+6m+1$, choose $y=0$. The order-zero bound gives $\lVert \Delta u^1\rVert\leq2M$, and the order-one bound gives $\lVert \Delta u^t\rVert\leq M\eta$ for $t\geq2$. Since $w\leq m/\eta$, \[ \sum_{t=1}^T\lVert \Delta u^t\rVert^2 \leq4M^2+TM^2\eta^2\leq(8m+5)M^2\leq D. \] The other conclusions are immediate.

For a longer horizon let $a=3m+1$, $b=T+1-3m$. Let $b_J$ be the probability mass $1/J$ on $\{0,\ldots,J-1\}$ and zero elsewhere, and let $q=b_J^{*m}$ be its $m$-fold discrete convolution. It is nonnegative, has sum one, and is supported on $\{0,\ldots,w\}$. Define \[ \chi=1_{\{a,\ldots,b-w\}}*q,\qquad y^t=\chi^t u^t\ (t\geq1),\qquad y^t=0\ (t\leq0). \] Then $0\leq\chi\leq1$, its support is contained in $[a,b]$, and $\chi=1$ on $[a+w,b-w]$. This proves (eq:cutoff-zero). Because $\lVert \Delta b_J\rVert_{\ell_1}=2/J$, convolution gives \[ \sup_t|\Delta^r\chi^t|\leq(2/J)^r\leq(2\eta)^r, \qquad 0\leq r\leq m. \] Let $\mathcal B$ be the integer starting indices whose stencil $\{t,\ldots,t+2m\}$ meets $[a,a+w]$ or $[b-w,b]$. Then \[ |\mathcal B|\leq2w+4m+2\leq8m/\eta, \qquad \min\mathcal B\geq a-2m=m+1. \] For $t\in\mathcal B$, all needed differences of $u$ therefore begin in the range where (eq:certificate-smooth) applies. The product rule \[ \Delta^j(\chi u)^s =\sum_{r=0}^j\binom jr(\Delta^r\chi)^s (\Delta^{j-r}u)^{s+r} \] implies $\lVert \Delta^j y^s\rVert\leq3^jM\eta^j$ on the stencils in question, for $0\leq j\leq m$.

Outside $\mathcal B$, a stencil has all cutoff values one or all zero. Use the original recurrence coefficients on the former stencils and zero coefficients everywhere else. The residual in the recurrence for $y$ is then supported in $\mathcal B$ and bounded pointwise by $3^mM\eta^m$. Its $\ell_2$ norm is at most $E\eta^{m-1/2}$. The triangle inequality and invariance of $\ell_2$ norms under integer shifts prove (eq:cutoff-high).

Finally, $\Delta y^t=\Delta u^t$ whenever $a+w\leq t\leq b-w-1$. At most $2w+6m+1\leq(8m+1)/\eta$ other indices $t\in\{2,\ldots,T\}$ remain. Their contribution is at most $(8m+1)M^2\eta$, and the initial difference contributes at most $4M^2$. This gives (eq:cutoff-one) with the stated $D$. ∎

5.2. Interpolation and numerical constants

For finite-support Hilbert-valued sequences, Parseval's identity and Hölder's inequality give \begin{equation}\label{wsref-0-65-71-3a-69-6e-74-65-72-70-6f-6c-61-74-69-6f-6e} U_j\leq U_0^{1-j/m}U_m^{j/m}\qquad(0\leq j\leq m). \end{equation} Indeed the Fourier multiplier for $\Delta$ has squared modulus $|e^{i\theta}-1|^2$; applying Hölder to its $j$th power proves the claim after integration. This proof also covers vector-valued sequences.

Combine (eq:cutoff-high) with (eq:interpolation). We claim \begin{equation}\label{wsref-0-65-71-3a-68-69-67-68-65-73-74-2d-65-78-70-6c-69-63-69-74} U_m\leq P^m\eta^m U_0+2E\eta^{m-1/2}. \end{equation} If $U_0=0$ all terms vanish. Otherwise, either $U_m\leq P^m\eta^m U_0$, or put $r=(U_m/(\eta^mU_0))^{1/m}>P\geq1$. In the latter case the sum on the right side of (eq:cutoff-high) is at most $AU_m\sum_{k<m}r^{k-m}\leq(Am/r)U_m\leq U_m/2$. Absorption proves (eq:highest-explicit) in either case.

For $m\geq1$ and nonnegative $z_1,z_2$, $(z_1+z_2)^{2/m}\leq2(z_1^{2/m}+z_2^{2/m})$. Using (eq:interpolation), (eq:highest-explicit), and weighted Young's inequality gives \begin{align*} U_1^2 &\leq2P^2\eta^2U_0^2 +2(2E)^{2/m}(\eta^2U_0^2)^{1-1/m}\eta^{1/m}\\ &\leq C\eta^2U_0^2+Q\eta, \end{align*} with the stated $Q$ and $C$. At $m=1$ this is read directly as $U_1^2\leq2P^2\eta^2U_0^2+Q\eta$; thus no ambiguous $0^0$ convention is needed. Lemma Lemma proves \begin{equation}\label{wsref-0-65-71-3a-74-65-6d-70-6f-72-61-6c-2d-65-78-70-6c-69-63-69-74} \sum_{t=1}^T\lVert \Delta h_i^t\rVert_2^2 \leq C\eta^2\sum_{t=1}^{T+1}\lVert h_i^t\rVert_2^2+D+Q\eta. \end{equation}

5.3. Completing the regret calculation

For completeness, the one-player variance inequalities used next are proved here with numerical constants. They are the variance form of the optimistic potential calculation used in Ha [ha]. Define $\Phi(z)=-(1/\eta)\log\sum_a e^{-\eta z_a}$ and let $D_{-\Phi}$ denote the Bregman divergence of the convex function $-\Phi$. If $p=\nabla\Phi(z)$ and the coordinate span of $w$ is at most $r$, its integral Hessian formula gives \[ \frac\eta2 e^{-\eta r}\operatorname{Var}_p(w) \leq D_{-\Phi}(z+w,z) \leq\frac\eta2 e^{\eta r}\operatorname{Var}_p(w). \] To check this, along the segment $z+sw$ each probability is between $e^{-\eta r}$ and $e^{\eta r}$ times its initial value. Apply these bounds to $\operatorname{Var}_p(w)=\min_c\sum_a p_a(w_a-c)^2$ and integrate $\eta(1-s)$ over $0\leq s\leq1$.

Let $L^t=\sum_{s\leq t}\ell^s$ and $q^t=L^{t-1}+\ell^{t-1}$. Then $x^t=\nabla\Phi(q^t)$, and direct expansion gives \[ D_{-\Phi}(L^t,q^t)-D_{-\Phi}(L^{t-1},q^t) =\langle x^t,\ell^t\rangle-\Phi(L^t)+\Phi(L^{t-1}). \] Sum this identity and compare $\Phi(L^T)$ with $\min_aL_a^T$. The coordinate spans of $\ell^t-\ell^{t-1}$ and $-\ell^{t-1}$ are at most two and one respectively. Thus, for $\eta\leq1/12$, \begin{equation}\label{wsref-0-65-71-3a-76-61-72-69-61-6e-63-65-2d-72-76-75} R_i(T)\leq\frac{\log m_i}{\eta} +\frac{2\eta}3\sum_{t=1}^T\operatorname{Var}_{x_i^t}(\ell_i^t-\ell_i^{t-1}) -\frac\eta3\sum_{t=1}^T\lVert h_i^t\rVert_2^2. \end{equation} Here $e^{2\eta}/2\leq2/3$ and $e^{-\eta}/2\geq1/3$ on the stated interval, which also follows from $e^z\leq1/(1-z)$ for $0\leq z<1$.

Write $w_{ab}^t=\sqrt{x_{ia}^tx_{ib}^t}$. The coordinate span of $2\ell_i^t-\ell_i^{t-1}$ is at most three, so $e^{-3\eta}\leq w_{ab}^{t+1}/w_{ab}^t\leq e^{3\eta}$. For $\eta\leq1/12$, $e^{3\eta}-1\leq4\eta$. Hence \begin{align*} &w_{ab}^t\left[(\ell_{ia}^t-\ell_{ib}^t) -(\ell_{ia}^{t-1}-\ell_{ib}^{t-1})\right]\\ &\qquad=(h_{i,ab}^{t+1}-h_{i,ab}^t) +\left(\frac{w_{ab}^t}{w_{ab}^{t+1}}-1\right)h_{i,ab}^{t+1}. \end{align*} Square with $(z+w)^2\leq(3/2)z^2+3w^2$, then sum over pairs and time. The pairwise variance identity and the terminal bound $\lVert h_i^{T+1}\rVert^2\leq1/4$ give \begin{equation}\label{wsref-0-65-71-3a-77-65-69-67-68-74-65-64-2d-67-61-70-2d-65-78-70-6c-69-63-69-74} \sum_{t=1}^T\operatorname{Var}_{x_i^t}(\ell_i^t-\ell_i^{t-1}) \leq\frac32\sum_{t=1}^T\lVert \Delta h_i^t\rVert_2^2 +48\eta^2\left(\sum_{t=1}^T\lVert h_i^t\rVert_2^2+\frac14\right). \end{equation} Insert (eq:temporal-explicit) and (eq:weighted-gap-explicit) into (eq:variance-rvu). Writing $H=\sum_{t=1}^T\lVert h_i^t\rVert^2$, we obtain \[ R_i(T)\leq\frac{\log m_i}{\eta}+\eta D+Q\eta^2 +\left((C+32)\eta^3-\frac\eta3\right)H +\frac{C+32}{4}\eta^3. \] The definition of $\bar\eta$ makes the coefficient of $H$ at most $-\eta/6$ and the last term at most $\eta/24$. Drop the nonpositive term and use $\eta\leq1$ on $Q\eta^2$. This proves Theorem Theorem (Quantitative recurrence criterion).

6. Conditional computability for all finite games

Here is a precise conditional way to make the outstanding recurrence search effective. Its hypothesis is a decision procedure for the full first-order theory of \[ \mathbb R_{\exp}=(\mathbb R;0,1,+,\cdot,<,\exp). \] Only sentences with rational coefficients are needed. This hypothesis is not supplied by real analyticity or by Noetherianity alone. The real form of Schanuel's conjecture implies such decidability; Berarducci and Gallinaro state the precise implication and give a new conditional axiomatization in [bg]. The theorem below needs only the decision procedure, not the stronger number-theoretic conjecture as a separate premise.
Theorem (Conditional effective search): Assume that the first-order theory of $\mathbb R_{\exp}$ is decidable. Then, using the qualitative finite-generation theorem and analytic state construction in Ha [ha, Theorem 5.5 and Lemmas 5.6--5.8], there is a terminating algorithm which, on input $(N,K)$, computes a positive common step size and a finite regret bound for the exact update (eq:update), uniformly over all allowed games and horizons. This is a conditional answer to the computability part of WS-107791. It does not imply the polynomial subquestion.
Proof: Fix an action-count vector $\mathbf d=(d_1,\ldots,d_N)$ with $1\leq d_i\leq K$. The parameter space of loss tables is $\mathcal G=[0,1]^{N\prod_j d_j}$. The real state space is the compact semialgebraic set \[ \mathcal S=\{(r,z): r_{ia}\geq0,\ \sum_a r_{ia}^2=1, \ 0\leq z_{ia}\leq1\text{ for all }i,a\}. \] Here $r_{ia}$ is a square-root probability and $z_i$ stores the preceding loss vector. For real $r$ near the product of unit spheres, set \[ p_{ia}(r)=\frac{r_{ia}^2}{\sum_b r_{ib}^2},\qquad L_{g,ia}(r)=\sum_{a_{-i}}g_i(a,a_{-i})\prod_{j\ne i}p_{j,a_j}(r). \] With $s=(r,z)$, define $G(s;g,\eta)=(F(s;g,\eta),L_g(r))$, where \[ F_{ia}(s;g,\eta)= \frac{\left(\sum_b r_{ib}^2\right)^{1/2}r_{ia} \exp\{-\eta(2L_{g,ia}(r)-z_{ia})/2\}} {\left(\sum_b r_{ib}^2 \exp\{-\eta(2L_{g,ib}(r)-z_{ib})\}\right)^{1/2}}. \] The table $g$ and step size $\eta$ remain fixed under composition of $G$. For real inputs in $\mathcal S$ every denominator is positive, and $G$ maps $\mathcal S$ into itself. The extra numerator norm is one on the state space; it ensures the ambient identity $G(r,z;g,0)=(r,L_g(r))$ and hence $G^2(s;g,0)=G(s;g,0)$.

Let $H_0(s)$ be the vector formed by stacking, for all players, the coordinates $r_{ia}r_{ib}(z_{ia}-z_{ib})$, $a<b$. For $\eta>0$, define the finite expression \begin{equation}\label{wsref-0-65-71-3a-65-78-70-6c-69-63-69-74-2d-48} H_k(s;g,\eta)=\eta^{-k} \sum_{j=0}^k(-1)^{k-j}\binom kj H_0\left(G^{k+j}(s;g,\eta)\right). \end{equation} Each $H_k$ uses at most $2k$ iterates and finitely many field operations, positive square roots, and exponentials. Along the actual trajectory $s^t=(\sqrt{x^t},\ell^{t-1})$, \[ H_k(s^t;g,\eta)=\eta^{-k}\Delta^k h^{t+k}, \] where $h$ stacks the players' weighted gaps. In particular the order-one expression controls $\Delta h^{t+1}$, not $\Delta h^t$ at $t=1$. The exceptional initial difference remains covered by the cutoff proof above.

For positive integers $(m,A,M,q)$ form the following first-order sentence: for every $s\in\mathcal S$, every $g\in\mathcal G$, and every $0<\eta\leq1/q$, \begin{equation}\label{wsref-0-65-71-3a-66-6f-2d-63-65-72-74-69-66-69-63-61-74-65} \begin{gathered} \exists c_0,\ldots,c_{m-1}\in\mathbb R: \quad |c_k|\leq A\ (k<m),\qquad H_m(s;g,\eta)=\sum_{k<m}c_k H_k(s;g,\eta),\\ \lVert H_k(s;g,\eta)\rVert_2^2\leq M^2 \qquad(0\leq k\leq m). \end{gathered} \end{equation} The universal quantifiers have the indicated antecedents; coefficients $c_k$ may depend on $s,g,\eta$. The norm squares and vector identities are finite sums and coordinate conjunctions. Replace every positive square root by an auxiliary variable $b>0$ with its defining equation $b^2=a$; replace divisions by their defining equations with nonzero denominators. This effectively converts (eq:fo-certificate) to a sentence in the language of $\mathbb R_{\exp}$ with rational coefficients. No quantification over functions or over infinitely many iterates occurs in a candidate sentence.

We next verify that at least one candidate is true. The compact set for the analytic argument is $\mathcal S\times\mathcal G\times\{0\}$, including the stored-loss coordinate independently of the current expected loss. Although (eq:explicit-H) displays division by $\eta$, each fixed $H_k$ extends analytically through zero. Indeed, equivalently, \[ H_{k+1}=\left(H_k\circ G^2-H_k\circ G\right)/\eta. \] The numerator vanishes at $\eta=0$ by the idempotence of $G(\cdot;g,0)$. Induction proves analytic divisibility, with a neighborhood allowed to shrink at each finite order. Ha's finite-generation theorem applies to the stacked vectors $H_0,H_1,\ldots$ just as it applies to an individual player's vector. It yields a finite $m$ and analytic scalar coefficients expressing $H_m$ in terms of $H_0,\ldots,H_{m-1}$ on a neighborhood of that compact set. Compactness then supplies a positive common step-size interval and finite upper bounds for the finitely many coefficients and $H_k$ involved. Choosing sufficiently large integers $A,M,q$ proves that (eq:fo-certificate) is true for at least one tuple. This existence argument does not itself compute the tuple.

Now enumerate all positive-integer tuples $(m,A,M,q)$ in increasing sum, breaking ties lexicographically. Apply the assumed decision procedure to each associated sentence. Each test terminates. Since a true candidate exists, the search eventually returns one. Its guarantees hold on the actual trajectory at every round by invariance of $\mathcal S$. The stacked norm bounds imply the corresponding bounds for each player's subvector, so Theorem Theorem (Quantitative recurrence criterion) applies with $\eta_0=1/q$.

There are only finitely many action-count vectors $\mathbf d\in\{1,\ldots,K\}^N$. Run this terminating search for each. For each vector compute $\bar\eta_{\mathbf d}$ and its constants $D_{\mathbf d},Q_{\mathbf d}$ from Theorem Theorem (Quantitative recurrence criterion). Choose \[ \eta(N,K)=\min_{\mathbf d}\bar\eta_{\mathbf d},\qquad B(N,K)=\frac{\log K}{\eta(N,K)} +\eta(N,K)\max_{\mathbf d} \left(D_{\mathbf d}+Q_{\mathbf d}+\frac1{24}\right). \] These are positive computable real quantities, and all their nonlogarithmic operations are algebraic operations on the returned integers. If a rational step size and integer bound are preferred, choose a smaller positive rational step size by algebraic comparison and an integer upper bound using certified rational approximations to $\log K$. Theorem Theorem (Quantitative recurrence criterion) proves the required horizon-uniform bound. ∎

7. What is proved and what is still missing

Theorems Theorem and Theorem are unconditional on their stated classes. Theorem Theorem (Quantitative recurrence criterion) is a proved implication with numerical constants, not an assumption-free solution of WS-107791. It reduces the outstanding quantitative task to producing, from $(N,K)$ alone, valid numbers $(m,A,M,\eta_0)$ in (eq:certificate-recurrence)--(eq:certificate-smooth), or an alternative argument of comparable force.

Ha supplies existence through an analytic module near the compact set of states and game tables at zero step size. In his notation the module generators are the analytic scaled differences $H_{i,k}$, and finite generation produces a relation $H_{i,m}=\sum_{k<m}c_kH_{i,k}$. Merely saying "enumerate possible parameters" does not provide a terminating verification procedure for that analytic identity. Section 6 supplies the missing verifier under the explicitly additional decidability hypothesis. Compactness alone does not compute a neighborhood radius, and no unconditional certificate-search algorithm is established here.

There is a second unresolved question: even an extremely large computable bound would not establish polynomial dependence on $N$ and $\log K$. Conversely, the static-payoff lower bound in Proposition Proposition constrains a proposed step size but does not exclude all polynomially suitable rules. The full answer and the polynomial subquestion therefore both remain UNRESOLVED unconditionally.

8. Checks performed

The proofs above retain the original initialization and common fixed step size; handle $m_i=1$; distinguish cumulative regret from its absolute value; and control the exceptional first difference separately in the cutoff construction. The bounds are independent of the horizon and, on their stated classes, of the payoff table. No numerical trajectory is used as proof. The primary Ha manuscript was inspected in its exact version 2609.22839v1; the source theorem and the explicitly stated quantitative gap match the catalogue formulation. The displayed special cases and reductions have been checked algebraically within this contribution, without a claim of formal verification or novelty.

Models used

OpenAI GPT-6 Astra Pro: mathematical derivation, adversarial checking, primary-source inspection, and drafting. Exact internal build identifier and formal reasoning configuration: unknown. The work was prepared by a parallel research agent within the same assistant session, followed by separate-agent review of the quantitative recurrence proof and conditional certificate search, followed by root review. No external referee approval is claimed.

Human contributors and roles

The user requested the problem-by-problem solution attempts and supplied the problem package. No independent human proof audit is claimed. The original problem suggestions are credited in the supplied package to Rida Laraki. The results cited below retain their original authors' credit.

References

  • [ha] Junsoo Ha. A Horizon-Independent Regret Bound for Optimistic Hedge in General-Sum Games. arXiv:2609.22839v1, 19 September 2026. Equation $5$, Theorem 4.1, text following Corollary 4.2, Sections 5.3--5.5, and Appendices B--D inspected. https://arxiv.org/html/2609.22839v1.
  • [rs] Alexander Rakhlin and Karthik Sridharan. Online Learning with Predictable Sequences. Proceedings of the 26th Annual Conference on Learning Theory, PMLR 30:993--1019, 2013. Official publication metadata and abstract checked; the identities needed here are proved in the text. https://proceedings.mlr.press/v30/Rakhlin13.html.
  • [tsuchiya] Taira Tsuchiya. Tight Regret Upper and Lower Bounds for Optimistic Hedge in Two-Player Zero-Sum Games. arXiv:2510.11691v1, 13 October 2025. Primary arXiv metadata and abstract checked; cited for the existing zero-sum literature, not as a proof dependency. https://arxiv.org/abs/2510.11691.
  • [bg] Alessandro Berarducci and Francesco Gallinaro. On the elementary theory of the real exponential field. arXiv:2603.08365v1, 9 March 2026. Abstract and Introduction inspected for the implication from the real form of Schanuel's conjecture to decidability, and for the distinction between model completeness and completeness of the restricted-exponential axioms. https://arxiv.org/html/2603.08365v1.

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