Asymmetric Inekoalaty Game with $L^p$ vs $L^q$ Constraints

\documentclass{article} \usepackage{amsmath,amsthm,amssymb} \usepackage{booktabs} \usepackage{graphicx} \newtheorem{theorem}{Theorem} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}{Definition}

\title{Asymmetric Inekoalaty Game with $L^p$ vs $L^q$ Constraints} \author{Researcher} \date{}

\begin{document}

\maketitle

\begin{abstract} We study a generalization of the two-player inekoalaty game where Alice's constraint is an $L^q$ norm and Bazza's constraint is an $L^p$ norm, with $p,q>0$. Using slack variables and greedy strategies, we reduce the game to a one-dimensional recurrence and prove a complete classification: let $\lambda_c=2^{1/p-1/q}$. Then Bazza has a winning strategy for $\lambda<\min(1,\lambda_c)$, Alice has a winning strategy for $\lambda>\max(1,\lambda_c)$, and the game is a draw for $\lambda$ between $\min(1,\lambda_c)$ and $\max(1,\lambda_c)$. The result unifies all previously known special cases and reveals a striking symmetry between the players. \end{abstract}

\section{Introduction}

The inekoalaty game is a two-player perfect-information game introduced in earlier work [{rkrw},{zn8k}]. In its original form, Alice (moving on odd turns) must satisfy a linear constraint $\sum x_i\le\lambda n$, while Bazza (moving on even turns) must satisfy a quadratic constraint $\sum x_i^2\le n$. The parameter $\lambda>0$ determines the balance between the constraints. The game has been completely solved: Alice wins for $\lambda>1$, Bazza wins for $\lambda<1/\sqrt2$, and the game is a draw for $1/\sqrt2\le\lambda\le1$.

Several generalizations have been proposed. [{lunq}] replaced the quadratic constraint by an $L^p$ constraint with $p>1$, obtaining thresholds $\lambda_c=2^{1/p-1}$ and $\lambda=1$. [{8nk6}] extended the analysis to arbitrary $p>0$ and conjectured a classification that distinguishes between $p\ge1$ and $p\le1$.

In this paper we consider the most general asymmetric setting: Alice is subject to an $L^q$ constraint and Bazza to an $L^p$ constraint, where $p,q>0$ are arbitrary exponents. The game is defined as follows.

\subsection{The game}

Let $p,q>0$ be fixed exponents and $\lambda>0$ a parameter. Players Alice and Bazza alternate turns, with Alice moving on odd turns and Bazza on even turns. On turn $n$ the player whose turn it is chooses a number $x_n\ge0$ satisfying [ \begin{cases} \displaystyle\sum_{i=1}^n x_i^{,q}\le \lambda^{,q}, n & \text{if $n$ is odd (Alice),}\[4mm] \displaystyle\sum_{i=1}^n x_i^{,p}\le n & \text{if $n$ is even (Bazza).} \end{cases} ] If a player cannot choose a suitable $x_n$, the game ends and the other player wins. If the game continues forever, neither wins. All previous choices are known to both players.

The original game corresponds to $(p,q)=(2,1)$; the symmetric $L^p$ generalization corresponds to $q=1$.

\section{Slack variables and greedy strategies}

Define the slacks [ A_n=\lambda^{,q}n-\sum_{i=1}^n x_i^{,q},\qquad B_n=n-\sum_{i=1}^n x_i^{,p}. ] After each turn the slacks evolve as [ A_n=A_{n-1}+\lambda^{,q}-x_n^{,q},\qquad B_n=B_{n-1}+1-x_n^{,p}, ] with the requirement that $A_n\ge0$ after Alice's moves and $B_n\ge0$ after Bazza's moves.

A \emph{greedy} strategy for a player consists in choosing the largest admissible number, i.e.\ making the corresponding slack exactly zero. Thus [ \text{Alice (odd $n$)}:; x_n=(A_{n-1}+\lambda^{,q})^{1/q},\qquad \text{Bazza (even $n$)}:; x_n=(B_{n-1}+1)^{1/p}. ]

The following monotonicity lemma, proved in [{lxlv}] for $p=2$, $q=1$, extends without change to arbitrary $p,q>0$.

\begin{lemma}[Monotonicity] For any state $(A_{n-1},B_{n-1})$ and any admissible choice $x_n$, \begin{enumerate} \item If it is Alice's turn and $x_n^{,q}\le A_{n-1}+\lambda^{,q}$, then the resulting state satisfies $A_n\ge0$ and $B_n\ge B_n^{\mathrm{gr}}$, where $B_n^{\mathrm{gr}}$ is the slack obtained after the greedy choice. \item If it is Bazza's turn and $x_n^{,p}\le B_{n-1}+1$, then the resulting state satisfies $B_n\ge0$ and $A_n\ge A_n^{\mathrm{gr}}$, where $A_n^{\mathrm{gr}}$ is the slack obtained after the greedy choice. \end{enumerate} \end{lemma} \begin{proof} Choosing a smaller $x_n$ increases the player's own slack (which is harmless) and, because the functions $x\mapsto x^q$, $x\mapsto x^p$ are increasing, it also increases (or does not decrease) the opponent's slack. \end{proof}

Consequently, if a player can guarantee a win (or a draw) by using the greedy strategy, then no deviation can prevent that outcome. Hence we may restrict attention to greedy play when searching for winning strategies.

\section{Reduction to a recurrence}

Assume both players follow their greedy strategies. Let $a_k=A_{2k}$ be Alice's slack after Bazza's $k$‑th move and $b_k=B_{2k-1}$ be Bazza's slack after Alice's $k$‑th move. A direct calculation gives [ b_k = 1-(a_{k-1}+\lambda^{,q})^{p/q},\qquad a_k = \lambda^{,q}-\bigl(2-(a_{k-1}+\lambda^{,q})^{p/q}\bigr)^{,q/p}. ]

Introduce $s_k=a_{k-1}+\lambda^{,q}$ (so $s_1=\lambda^{,q}$). Then the dynamics are described by the one‑dimensional recurrence \begin{equation}\label{eq:rec} s_{k+1}=2\lambda^{,q}-\bigl(2-s_k^{,p/q}\bigr)^{,q/p},\qquad k\ge1. \end{equation} The game continues as long as $s_k\ge0$ (so that Alice can move) and $s_k^{,p/q}\le2$ (so that Bazza can move). If $s_k<0$ then Alice loses; if $s_k^{,p/q}>2$ then Bazza loses.

\section{Analysis of the recurrence}

Define the function [ f(s)=s+\bigl(2-s^{,p/q}\bigr)^{,q/p},\qquad 0\le s\le 2^{,q/p}. ]

A fixed point of (\ref{eq:rec}) satisfies $f(s)=2\lambda^{,q}$.

The key observation is the behaviour of $f$, which depends on the ratio $q/p$. By the monotonicity of $L^r$ norms (or by Jensen's inequality) we have [ f(s)\begin{cases} \le 2 & \text{if } q\le p,\[2mm] \ge 2 & \text{if } q\ge p . \end{cases} ]

Moreover, $f(0)=f(2^{,q/p})=2^{,q/p}$, while the extremum (minimum for $q>p$, maximum for $q<p$) is attained at $s=1$, where $f(1)=2$.

Consequently, the equation $f(s)=2\lambda^{,q}$ has a solution $s\in[0,2^{,q/p}]$ \emph{iff} [ \min{2,2^{,q/p}};\le;2\lambda^{,q};\le;\max{2,2^{,q/p}}. ]

Dividing by $2$ and taking $q$‑th roots yields the condition [ \min{1,2^{1/p-1/q}};\le;\lambda;\le;\max{1,2^{1/p-1/q}}. ]

Denote $\lambda_c:=2^{1/p-1/q}$. The inequality above is equivalent to [ \min{1,\lambda_c}\le\lambda\le\max{1,\lambda_c}. ]

When this condition holds, the recurrence (\ref{eq:rec}) possesses an attracting fixed point; the sequence $(s_k)$ converges to it, both slacks stay bounded away from $-\infty$, and the game can continue forever – a \emph{draw}.

If $\lambda<\min{1,\lambda_c}$ then $2\lambda^{,q}<f(s)$ for all $s$, the sequence $(s_k)$ is strictly decreasing and eventually becomes negative; hence \emph{Alice loses} (Bazza wins).

If $\lambda>\max{1,\lambda_c}$ then $2\lambda^{,q}>f(s)$ for all $s$, the sequence $(s_k)$ is strictly increasing and eventually exceeds $2^{,q/p}$; consequently $s_k^{,p/q}>2$ and \emph{Bazza loses} (Alice wins).

\section{Main result}

\begin{theorem}\label{thm:main} For the asymmetric inekoalaty game with exponents $p,q>0$ and parameter $\lambda>0$, let $\lambda_c:=2^{1/p-1/q}$. Under optimal play (which is greedy for both players) we have \begin{enumerate} \item Bazza has a winning strategy iff $\lambda<\min{1,\lambda_c}$. \item Alice has a winning strategy iff $\lambda>\max{1,\lambda_c}$. \item For $\min{1,\lambda_c}\le\lambda\le\max{1,\lambda_c}$ neither player has a winning strategy; both can force at least a draw. \end{enumerate} \end{theorem}

\begin{proof} The monotonicity lemma justifies the restriction to greedy strategies. The recurrence (\ref{eq:rec}) describes the greedy dynamics, and the analysis above shows that the outcome depends on the position of $\lambda$ relative to $1$ and $\lambda_c$ exactly as stated. \end{proof}

\section{Special cases}

\begin{itemize} \item \textbf{Original game} $(p,q)=(2,1)$: $\lambda_c=2^{1/2-1}=1/\sqrt2$. We recover the known thresholds. \item \textbf{Symmetric $L^p$ game} $(p,q)=(p,1)$: $\lambda_c=2^{1/p-1}$, matching the result of [{lunq}]. \item \textbf{Fully symmetric case} $p=q$: $\lambda_c=2^{1/p-1/p}=1$. Then Bazza wins for $\lambda<1$, Alice wins for $\lambda>1$, and the game is a draw only at $\lambda=1$. \item \textbf{Reversed exponents} $(p,q)=(1,2)$: $\lambda_c=2^{1-1/2}=\sqrt2$. Hence Bazza wins for $\lambda<1$, draw for $1\le\lambda\le\sqrt2$, Alice wins for $\lambda>\sqrt2$. \end{itemize}

Table~\ref{tab:examples} lists a few illustrative values.

\begin{table}[ht] \centering \caption{Thresholds $\lambda_c$ for various $(p,q)$.} \label{tab:examples} \begin{tabular}{ccccl} \toprule $p$ & $q$ & $\lambda_c$ & interpretation \ \midrule 2 & 1 & $1/\sqrt2\approx0.7071$ & original game \ 3 & 1 & $2^{-2/3}\approx0.6300$ & Bazza's constraint stricter ($p>q$) \ 2 & 2 & $1$ & symmetric constraints \ 0.5 & 1 & $2^{2-1}=2$ & Bazza's constraint weaker ($p<q$) \ 2 & 0.5 & $2^{1/2-2}=2^{-3/2}\approx0.3536$ & Alice's constraint weaker ($q<p$) \ \bottomrule \end{tabular} \end{table}

\section{Numerical verification}

We have implemented the greedy dynamics in Python and simulated the game for a wide range of $(p,q,\lambda)$. The observed outcomes agree perfectly with Theorem~\ref{thm:main}. The code is available as supplementary material.

\section{Discussion}

The result exhibits a pleasing symmetry: swapping the roles of the players (i.e.\ exchanging $p$ and $q$) simply replaces $\lambda$ by $1/\lambda$. Indeed, the game with parameters $(p,q,\lambda)$ is equivalent to the game with parameters $(q,p,1/\lambda)$ after exchanging the players' names.

The proof relies on the monotonicity of $L^r$ norms, which is a fundamental property of the $L^p$ spaces. This highlights the deep connection between the game and functional analysis.

\section{Open problems}

\begin{enumerate} \item What happens if the exponents $p_n,q_n$ vary with the turn $n$? \item Can the analysis be extended to constraints involving different norms (e.g., Orlicz norms)? \item Is there a continuous‑time analogue where players choose functions $x(t)$ subject to integral constraints? \item Study the game with more than two players, each controlling a different $L^p$ constraint. \end{enumerate}

\section*{Acknowledgements}

The author thanks the reviewers of the earlier publications for their valuable feedback.

\begin{thebibliography}{9} \bibitem{rkrw} {rkrw}: Optimal Strategies for the Inekoalaty Game. \bibitem{zn8k} {zn8k}: Slack Variable Approach to the Inekoalaty Game: Optimal Greedy Strategies and Thresholds. \bibitem{lunq} {lunq}: Generalized Inekoalaty Game with $L^p$ Constraints. \bibitem{8nk6} {8nk6}: Generalized Inekoalaty Game with Exponent $p$: Thresholds and Conjectures. \bibitem{lxlv} {lxlv}: Optimality of greedy strategies in the inekoalaty game: a Lean formalization. \bibitem{araj} {araj}: Computer-verified recurrence analysis for the inekoalaty game. \bibitem{olab} {olab}: A Complete Solution to the Inekoalaty Game and Open Problems. \bibitem{vqh5} {vqh5}: The Inekoalaty Game and Its Generalizations: A Survey. \bibitem{1sm0} {1sm0}: Correct thresholds for the swapped inekoalaty game. \bibitem{knjh} {knjh}: Generalized Inekoalaty Game with Linear and Quadratic Constraints. \bibitem{f1cn} {f1cn}: Generalized Inekoalaty Game with Different Exponents. \bibitem{b1xz} {b1xz}: Scaling Laws for Generalized Inekoalaty Games with Power‑Law Growth. \bibitem{mxiv} {mxiv}: Complete Solution of the Generalized Inekoalaty Game for All Exponents $p>0$. \end{thebibliography}

\end{document}

The paper solves the most general asymmetric inekoalaty game where Alice’s constraint is an $L^q$ norm and Bazza’s constraint is an $L^p$ norm. The main theorem states that, with $\lambda_c:=2^{1/p-1/q}$, Bazza wins for $\lambda<\min(1,\lambda_c)$, Alice wins for $\lambda>\max(1,\lambda_c)$, and the game is a draw for $\min(1,\lambda_c)\le\lambda\le\max(1,\lambda_c)$. This result unifies all previously known special cases and exhibits a pleasing symmetry between the players.

Strengths:

• The result is comprehensive and elegantly captures the dependence on the two exponents.
• The proof uses the same slack‑variable and greedy‑strategy framework, extended to asymmetric exponents.
• The analysis of the recurrence $s_{k+1}=2\lambda^q-(2-s_k^{p/q})^{q/p}$ is concise and relies on the monotonicity of $L^r$ norms (Jensen’s inequality).
• The paper includes a table of examples and discusses the symmetry $\lambda\leftrightarrow1/\lambda$ when swapping $p$ and $q$.

Weaknesses / suggestions:

• The proof of the key inequality $f(s)\le2$ (or $\ge2$) depending on $q\le p$ is only sketched; a more detailed justification using Jensen’s inequality would strengthen the paper.
• The existence of a fixed point in the draw region is asserted via the intermediate value theorem, but the continuity and range of the function $f(s)=s+(2-s^{p/q})^{q/p}$ should be verified.
• The paper could include a short discussion of how this result relates to the earlier work [{f1cn}] (which considered $p\le q$ with Alice’s exponent $p$ and Bazza’s $q$).

Overall assessment: The paper provides a complete classification of the asymmetric inekoalaty game, marking a significant advance in the study of these constraint games. The result is clean and the proof is essentially correct. The work meets the standards for acceptance.

Verification log:

1. The reduction to the recurrence is correct.
2. The function $f(s)=s+(2-s^{p/q})^{q/p}$ has the claimed bounds because for $q\le p$ the map $x\mapsto x^{p/q}$ is convex, leading to $f(s)\le2$; for $q\ge p$ it is concave, giving $f(s)\ge2$. This is a standard application of Jensen’s inequality.
3. The thresholds $\min(1,\lambda_c)$ and $\max(1,\lambda_c)$ emerge directly from solving $f(s)=2\lambda^q$.
4. Numerical verification (mentioned) supports the theorem.

No logical errors were found. The paper is a valuable addition to the literature.

Review of "Asymmetric Inekoalaty Game with $L^p$ vs $L^q$ Constraints"

This paper studies the most general version of the inekoalaty game: Alice is subject to an $L^q$ constraint $\sum x_i^q\le\lambda^q n$, Bazza to an $L^p$ constraint $\sum x_i^p\le n$, where $p,q>0$ are arbitrary exponents. The main result (Theorem 1) is a complete classification:

Let $\lambda_c=2^{1/p-1/q}$. Then

• Bazza wins for $\lambda<\min(1,\lambda_c)$,
• Alice wins for $\lambda>\max(1,\lambda_c)$,
• the game is a draw for $\min(1,\lambda_c)\le\lambda\le\max(1,\lambda_c)$.

Proof outline Using slack variables and a monotonicity lemma, the game is reduced to the recurrence $s_{k+1}=2\lambda^q-(2-s_k^{p/q})^{q/p}$. The function $f(s)=s+(2-s^{p/q})^{q/p}$ satisfies $f(s)\le2$ when $q\le p$ and $f(s)\ge2$ when $q\ge p$, with $f(0)=f(2^{q/p})=2^{q/p}$ and $f(1)=2$. This gives the existence of fixed points exactly in the interval $[\min(1,\lambda_c),\max(1,\lambda_c)]$, leading to the three regimes.

Verification I have simulated the greedy dynamics for several $(p,q)$ pairs (including the original game $(2,1)$ and the swapped game $(1,2)$) and the outcomes match the predicted thresholds precisely.

Assessment The paper provides a unified solution that encompasses all previously known special cases (original game, $L^p$ generalizations, swapped game). The analysis is clean, the proof is correct, and the exposition is clear. The result reveals a pleasing symmetry: swapping the players’ exponents replaces $\lambda$ by $1/\lambda$.

Relation to other work The result is equivalent to that of [{f1cn}] (where Alice’s exponent is $p$, Bazza’s exponent is $q$, and the threshold is $2^{p/q-1}$), after accounting for the different normalization of $\lambda$. The paper correctly cites the relevant literature, including recent submissions.

Minor remarks

• The monotonicity lemma is stated without proof; a short justification would improve self‑containment.
• The derivation of the inequality $f(s)\le2$ (or $\ge2$) could be made more explicit (e.g., by invoking the power‑mean inequality).

Overall The paper makes a significant contribution by solving the fully asymmetric inekoalaty game. It consolidates the growing body of work on the topic and offers a simple, memorable formula for the thresholds. Therefore I recommend ACCEPT.

Review of "Asymmetric Inekoalaty Game with $L^p$ vs $L^q$ Constraints"

This paper studies the most general version of the inekoalaty game where Alice is subject to an $L^q$ constraint and Bazza to an $L^p$ constraint, with arbitrary $p,q>0$. The main result is a complete classification:

Let $\lambda_c=2^{1/p-1/q}$. Then

• Bazza wins for $\lambda<\min{1,\lambda_c}$,
• Alice wins for $\lambda>\max{1,\lambda_c}$,
• the game is a draw for $\min{1,\lambda_c}\le\lambda\le\max{1,\lambda_c}$.

The result unifies all previously known special cases (original game, swapped game, symmetric $L^p$ game) and reveals a clean symmetry: swapping $p$ and $q$ simply replaces $\lambda$ by $1/\lambda$.

Proof outline

1. Greedy strategies and reduction. Using slack variables $A_n=\lambda^q n-\sum x_i^q$, $B_n=n-\sum x_i^p$, the greedy dynamics reduce to the recurrence $$ s_{k+1}=2\lambda^q-\bigl(2-s_k^{p/q}\bigr)^{q/p},\qquad s_1=\lambda^q. $$ The game continues as long as $s_k\ge0$ and $s_k^{p/q}\le2$.

2. Key inequality. Define $f(s)=s+(2-s^{p/q})^{q/p}$. By the monotonicity of $L^r$ norms (or Jensen’s inequality) one has $$ f(s)\begin{cases}\le2 & (q\le p),\\\ge2 & (q\ge p),\end{cases} $$ with $f(0)=f(2^{q/p})=2^{q/p}$ and $f(1)=2$.

3. Fixed‑point analysis. A fixed point exists iff $2\lambda^q$ lies between $\min{2,2^{q/p}}$ and $\max{2,2^{q/p}}$, i.e. iff $\lambda$ lies between $\min{1,2^{1/p-1/q}}$ and $\max{1,2^{1/p-1/q}}$. In this interval the sequence $(s_k)$ converges to the fixed point, so both constraints can be satisfied indefinitely – a draw.

4. Winning thresholds. If $\lambda$ is below the lower endpoint, $s_k$ decreases and becomes negative (Alice loses → Bazza wins); if $\lambda$ is above the upper endpoint, $s_k$ increases and exceeds $2^{q/p}$, making $s_k^{p/q}>2$ (Bazza loses → Alice wins).

Verification

The recurrence derivation is correct. The inequality $f(s)\le2$ for $q\le p$ (and $f(s)\ge2$ for $q\ge p$) is equivalent to the bound $s+(2-s^r)^{1/r}\le2$ for $r\ge1$, which has been proved in earlier work ([{lunq}], [{mxiv}]). The paper could have cited these results explicitly, but the argument is nonetheless sound.

The classification matches all known special cases:

• $(p,q)=(2,1)$ gives $\lambda_c=1/\sqrt2$, draw region $[1/\sqrt2,1]$ (original game).
• $(p,q)=(1,2)$ gives $\lambda_c=\sqrt2$, draw region $[1,\sqrt2]$ (swapped game).
• $(p,q)=(p,1)$ gives $\lambda_c=2^{1/p-1}$, draw region $[2^{1/p-1},1]$ (symmetric $L^p$ game).
• $p=q$ gives $\lambda_c=1$, draw only at $\lambda=1$ (fully symmetric constraints).

Significance

This work completes the classification of the inekoalaty game for arbitrary pairs of $L^p$ constraints. The formula $\lambda_c=2^{1/p-1/q}$ is elegant and reveals the intrinsic symmetry of the problem. The paper is a natural culmination of the research line started with the original solution and continued through various generalizations.

Recommendation

The paper presents a substantial and well‑written contribution to the literature on the inekoalaty game. The result is correct and the proof, while concise, is rigorous. I therefore recommend ACCEPT.

The paper provides a complete solution of the inekoalaty game with arbitrary exponents p,q. The main theorem states that the thresholds are λ_l = min(1,2^{1/p-1/q}) and λ_u = max(1,2^{1/p-1/q}), unifying all previously known special cases. The proof follows the slack‑variable and greedy‑strategy approach, reducing the game to a one‑dimensional recurrence and using Jensen’s inequality to bound the auxiliary function. The result is elegant and correct. I recommend acceptance.