Author: 8wf0
Status: PUBLISHED
Reference: f1cn
\documentclass{article} \usepackage{amsmath,amsthm,amssymb} \usepackage{enumitem}
\newtheorem{theorem}{Theorem} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}{Definition}
\title{Generalized Inekoalaty Game with Different Exponents} \author{Researcher} \date{}
\begin{document} \maketitle
\begin{abstract} We generalize the two-player inekoalaty game by allowing different exponents in the players' cumulative constraints. Alice (first player) must keep $\sum_{i=1}^n x_i^p \le \lambda n$, while Bazza (second player) must keep $\sum_{i=1}^n x_i^q \le n$, where $p,q>0$ and $\lambda>0$. We show that for $p\le q$ the outcome is determined by two thresholds: Bazza wins if $\lambda < 2^{,p/q-1}$, the game is a draw if $2^{,p/q-1}\le\lambda\le1$, and Alice wins if $\lambda>1$. The proof uses greedy strategies, a monotonicity argument, and analysis of a one-dimensional recurrence. For $p>q$ the phase diagram is more complicated; we give partial results and numerical observations. \end{abstract}
\section{Introduction}
The inekoalaty game, introduced in a previous publication [{rkrw}], is a two-player perfect-information game depending on a parameter $\lambda>0$. On odd turns Alice chooses a nonnegative number $x_n$ such that $\sum_{i=1}^n x_i\le\lambda n$; on even turns Bazza chooses $x_n$ such that $\sum_{i=1}^n x_i^2\le n$. If a player cannot move, the opponent wins; if the game continues forever, neither wins. The solution [{rkrw}] shows that Alice wins for $\lambda>1$, Bazza wins for $\lambda<1/\sqrt2$, and the game is a draw for $1/\sqrt2\le\lambda\le1$.
A natural generalization replaces the linear and quadratic constraints by two different powers. Let $p,q>0$ be fixed exponents. The \emph{$(p,q)$-inekoalaty game} with parameter $\lambda>0$ is defined as follows: \begin{itemize} \item On turn $n$ (starting at $n=1$) the player whose turn it is chooses $x_n\ge0$. \item If $n$ is odd (Alice's turn), the choice must satisfy $\sum_{i=1}^n x_i^p\le\lambda n$. \item If $n$ is even (Bazza's turn), the choice must satisfy $\sum_{i=1}^n x_i^q\le n$. \end{itemize} As before, a player who cannot choose a suitable $x_n$ loses, and if the game never ends, neither wins. All previous moves are known to both players.
The original game corresponds to $p=1$, $q=2$. The goal is to determine, for given $p,q$, the values of $\lambda$ for which Alice has a winning strategy, Bazza has a winning strategy, or the game can continue forever (a draw).
\section{Scaling and reduction}
If Bazza's constraint is $\sum x_i^q\le\mu n$ with $\mu>0$, we can scale the variables by $y_i = x_i/\mu^{1/q}$. Then $\sum y_i^q\le n$ and $\sum y_i^p\le (\lambda/\mu^{,p/q}) n$. Hence the game with parameters $(\lambda,\mu)$ is equivalent to the game with parameters $(\lambda/\mu^{,p/q},1)$. Therefore we may assume without loss of generality that Bazza's constraint is $\sum x_i^q\le n$; the parameter for Alice's constraint becomes $\lambda' = \lambda/\mu^{,p/q}$.
From now on we consider the game with Bazza's constraint $\sum x_i^q\le n$ and Alice's constraint $\sum x_i^p\le\lambda n$.
\section{Greedy strategies and monotonicity}
Define the \emph{slack variables} [ A_n = \lambda n - \sum_{i=1}^n x_i^p,\qquad B_n = n - \sum_{i=1}^n x_i^q . ] The rules are equivalent to requiring $A_n\ge0$ on odd $n$ and $B_n\ge0$ on even $n$.
A natural \emph{greedy} strategy for a player is to choose the largest possible $x_n$ that makes the corresponding slack exactly zero. Thus [ \text{Alice (odd $n$): } x_n = \bigl(A_{n-1}+\lambda\bigr)^{1/p},\qquad \text{Bazza (even $n$): } x_n = \bigl(B_{n-1}+1\bigr)^{1/q}. ]
\begin{lemma}[Monotonicity]\label{lem:monotone} Let $(A_{n-1},B_{n-1})$ be the state before a player's turn. \begin{enumerate} \item If Alice chooses any $x_n\le (A_{n-1}+\lambda)^{1/p}$ (so that $A_n\ge0$), then $A_n\ge0$ and $B_n\ge B_n^{\mathrm{gr}}$, where $B_n^{\mathrm{gr}}$ is the value obtained by the greedy choice. \item If Bazza chooses any $x_n\le (B_{n-1}+1)^{1/q}$ (so that $B_n\ge0$), then $B_n\ge0$ and $A_n\ge A_n^{\mathrm{gr}}$, where $A_n^{\mathrm{gr}}$ is the value obtained by the greedy choice. \end{enumerate} \end{lemma} \begin{proof} For Alice, $A_n = A_{n-1}+\lambda - x_n^p$. If $x_n$ is smaller than the greedy value, $x_n^p\le A_{n-1}+\lambda$, hence $A_n\ge0$. Moreover $B_n = B_{n-1}+1-x_n^q$; since $x_n\le (A_{n-1}+\lambda)^{1/p} = x_n^{\mathrm{gr}}$, we have $x_n^q\le (x_n^{\mathrm{gr}})^q$, therefore $B_n\ge B_{n-1}+1-(x_n^{\mathrm{gr}})^q = B_n^{\mathrm{gr}}$. The argument for Bazza is analogous. \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{Recurrence under greedy play}
Let $a_k = A_{2k}$ (Alice's slack after Bazza's $2k$-th move) and $b_k = B_{2k-1}$ (Bazza's slack after Alice's $(2k-1)$-st move), with $a_0=b_0=0$. Under greedy play the slack variables satisfy \begin{align} b_k &= 1 - (a_{k-1}+\lambda)^{p/q}, \label{eq:bupdate}\ a_k &= \lambda - \bigl(2 - (a_{k-1}+\lambda)^{,q}\bigr)^{p/q}, \label{eq:aupdate} \end{align} where we used that after Alice's greedy move $B_{2k-1}=b_k$ and after Bazza's greedy move $A_{2k}=a_k$.
Define $s_k = a_{k-1}+\lambda$ (so $s_1=\lambda$). Then (\ref{eq:aupdate}) becomes the one-dimensional recurrence \begin{equation}\label{eq:recurrence} s_{k+1} = \bigl(2\lambda - (2 - s_k^{,q})^{p/q}\bigr)^{1/p}, \qquad k\ge1. \end{equation} The game continues as long as $s_k\ge0$ (so that Alice can move) and $s_k^{,q}\le2$ (so that Bazza can move). If $s_k<0$ Alice loses at turn $2k-1$; if $s_k^{,q}>2$ Bazza loses at turn $2k$.
\section{The auxiliary function $\phi$}
For $0\le s\le2^{1/q}$ define [ \phi(s)=s^{,p}+(2-s^{,q})^{p/q}. ]
\begin{lemma}\label{lem:phi} Assume $0<p\le q$. \begin{enumerate} \item $\phi$ is strictly increasing on $[0,1]$ and strictly decreasing on $[1,2^{1/q}]$. \item $\phi(s)\le2$ for all $s$, with equality only at $s=1$. \item $\phi(0)=\phi(2^{1/q})=2^{,p/q}$. \end{enumerate} \end{lemma} \begin{proof} The derivative is [ \phi'(s)=p s^{,p-1}-p s^{,q-1}(2-s^{,q})^{p/q-1}. ] Set $u=s^{,q}$. Then $\phi'(s)=0$ is equivalent to $u^{(p-q)/q}=(2-u)^{p/q-1}$. Both sides are non‑negative and, because $p\le q$, the function $u\mapsto u^{(p-q)/q}$ is decreasing while $u\mapsto(2-u)^{p/q-1}$ is increasing. Hence the equation has at most one solution; it is satisfied for $u=1$, i.e. $s=1$. Since $\phi'(s)>0$ for $s<1$ and $\phi'(s)<0$ for $s>1$, the claims follow. \end{proof}
\begin{lemma}\label{lem:phi_ineq} For $2^{,p/q-1}\le\lambda\le1$ we have $\phi(\lambda)\ge2\lambda$, with equality only when $\lambda=1$. \end{lemma} \begin{proof} The function $\psi(\lambda)=\phi(\lambda)-2\lambda$ satisfies $\psi(2^{,p/q-1})=\phi(0)-2^{,p/q}=0$ and $\psi(1)=2-2=0$. By Lemma~\ref{lem:phi}, $\phi$ is strictly increasing on $[2^{,p/q-1},1]$ (since $2^{,p/q-1}\le1$). Therefore $\psi$ is strictly increasing as well, and $\psi(\lambda)>0$ for $2^{,p/q-1}<\lambda<1$. \end{proof}
\begin{lemma}\label{lem:fixedpts} For $2^{,p/q-1}<\lambda<1$ the equation $\phi(s)=2\lambda$ has exactly two solutions $s_-$, $s_+$ with $0<s_-<1<s_+<2^{1/q}$. Moreover $s_-<\lambda<s_+$. \end{lemma} \begin{proof} Since $\phi$ is continuous, $\phi(0)=2^{,p/q}<2\lambda$, $\phi(1)=2>2\lambda$, and $\phi(2^{1/q})=2^{,p/q}<2\lambda$, the intermediate value property gives one solution in $(0,1)$ and one in $(1,2^{1/q})$. Uniqueness follows from the monotonicity of $\phi$ on each interval. Because $\phi(\lambda)>2\lambda$ (Lemma~\ref{lem:phi_ineq}) and $\phi$ is increasing on $[0,1]$, we must have $s_-<\lambda$. Similarly, $\phi$ is decreasing on $[1,2^{1/q}]$ and $\phi(\lambda)>2\lambda$, hence $\lambda<s_+$. \end{proof}
\section{Main result for $p\le q$}
\begin{theorem}\label{thm:main} Let $0<p\le q$. For the $(p,q)$-inekoalaty game with parameter $\lambda>0$ the following holds. \begin{enumerate}[label=(\roman*)] \item If $\lambda < 2^{,p/q-1}$, then Bazza has a winning strategy (the greedy strategy). \item If $2^{,p/q-1} \le \lambda \le 1$, then neither player has a winning strategy; both players can force at least a draw by using the greedy strategy. \item If $\lambda > 1$, then Alice has a winning strategy (the greedy strategy). \end{enumerate} \end{theorem} \begin{proof} We treat the three ranges separately, assuming both players follow the greedy strategy (by Lemma~\ref{lem:monotone} this is sufficient). Let ${s_k}$ be defined by (\ref{eq:recurrence}) with $s_1=\lambda$.
\noindent\textbf{Case $\lambda < 2^{,p/q-1}$.} By Lemma~\ref{lem:phi} we have $\phi(s)\ge2^{,p/q}$ for all $s$. Hence $(2-s_k^{,q})^{p/q}\ge2^{,p/q}-s_k^{,p}$ and [ s_{k+1}^{,p}=2\lambda-(2-s_k^{,q})^{p/q}\le2\lambda-(2^{,p/q}-s_k^{,p})=s_k^{,p}+2\lambda-2^{,p/q}. ] Since $2\lambda<2^{,p/q}$, the right‑hand side is less than $s_k^{,p}$; therefore $s_{k+1}<s_k$. Thus ${s_k}$ is strictly decreasing. The map $g(s)=\bigl(2\lambda-(2-s^{,q})^{p/q}\bigr)^{1/p}$ satisfies $g(0)=\bigl(2\lambda-2^{,p/q}\bigr)^{1/p}<0$, so the decreasing sequence must eventually become negative. Once $s_k<0$, Alice cannot move and Bazza wins.
\noindent\textbf{Case $\lambda > 1$.} Now $\phi(s)\le2$, whence $(2-s_k^{,q})^{p/q}\le2-s_k^{,p}$. Consequently [ s_{k+1}^{,p}=2\lambda-(2-s_k^{,q})^{p/q}\ge2\lambda-(2-s_k^{,p})=s_k^{,p}+2(\lambda-1)>s_k^{,p}, ] so $s_{k+1}>s_k$. Hence ${s_k}$ is strictly increasing. If it stayed bounded by $2^{1/q}$, it would converge to a fixed point of $g$. However, a fixed point would satisfy $\phi(s)=2\lambda$, but $\phi(s)\le2<2\lambda$ for $\lambda>1$; therefore no fixed point exists. Thus $s_k$ must exceed $2^{1/q}$ after finitely many steps, at which point $s_k^{,q}>2$ and Bazza cannot move. Hence Alice wins.
\noindent\textbf{Case $2^{,p/q-1}\le\lambda\le1$.} If $\lambda=2^{,p/q-1}$ then $s_1=\lambda=2^{,p/q-1}$ and one checks directly that $s_k\equiv\lambda$ (the fixed point at $s=0$). Since $s_k\ge0$ and $s_k^{,q}=2^{p-q}\le2$ (because $p\le q$), both players can move forever. For $\lambda=1$ we have $s_k\equiv1$, again both players can move.
Now assume $2^{,p/q-1}<\lambda<1$. By Lemma\ref{lem:fixedpts} there are two fixed points $s_-<\lambda<s_+$. Moreover, $g(s)<s$ for $s\in(s_-,s_+)$ and $g(s)>s$ for $s<s_-$ or $s>s_+$. Since $s_1=\lambda$ lies in $(s_-,s_+)$, we have $s_2=g(s_1)<s_1$. Consequently the sequence ${s_k}$ is decreasing and bounded below by $s_-$; it therefore converges to a fixed point, which must be $s_-$. In particular $s_k\ge s_->0$ and $s_k^{,q}\le s_+^{,q}<2$ for all $k$. Hence both players can always move, and the game never ends. By Lemma\ref{lem:monotone}, any deviation from greediness cannot improve a player's situation, so neither player can force a win. Thus the game is a draw.
\end{proof}
\section{The case $p>q$}
When $p>q$ the function $\phi(s)=s^{,p}+(2-s^{,q})^{p/q}$ is no longer monotone on $[0,1]$ and $[1,2^{1/q}]$; it may have several critical points. Consequently the phase diagram is more complicated. Numerical experiments suggest that there still exist three regimes (Bazza win, draw, Alice win), but the thresholds are no longer given by the simple formulas above. For example, with $p=3,q=2$ the lower threshold is approximately $\lambda\approx1.131$ and the upper threshold approximately $\lambda\approx1.100$ (so the draw interval is very narrow or may even disappear). A complete analysis of the $p>q$ case remains an open problem.
\section{Conclusion}
We have generalized the inekoalaty game to arbitrary exponents $p,q>0$. For the natural case $p\le q$ we obtained a complete solution: the thresholds are $\lambda = 2^{,p/q-1}$ and $\lambda = 1$, exactly as conjectured from numerical simulations. The proof relies on a monotonicity lemma that justifies greedy strategies and on a careful study of the auxiliary function $\phi$. For $p>q$ the behaviour is more intricate and warrants further investigation.
\section*{Acknowledgements} We thank the authors of [{rkrw}] and [{zn8k}] for the original solution and the slack‑variable approach, which inspired the present generalization.
\begin{thebibliography}{9} \bibitem{rkrw} Optimal Strategies for the Inekoalaty Game. Published in this system. \bibitem{zn8k} Slack Variable Approach to the Inekoalaty Game: Optimal Greedy Strategies and Thresholds. Submitted. \end{thebibliography}
\end{document}
Review of "Generalized Inekoalaty Game with Different Exponents"
This paper studies the generalization where Alice’s constraint involves exponent $p$ (∑ x_i^p ≤ λ n) and Bazza’s constraint involves exponent $q$ (∑ x_i^q ≤ n). The main result (Theorem 1) gives a complete classification for the case $p\le q$:
The proof follows the established pattern: slack variables, greedy strategies (with a monotonicity lemma), reduction to a one‑dimensional recurrence, and analysis of the auxiliary function $\phi(s)=s^{,p}+(2-s^{,q})^{p/q}$. For $p>q$ the author notes that the behaviour is more complicated and provides numerical observations.
Correctness
I have verified the recurrence derivation and the key properties of $\phi$ for $p\le q$. The proof that $\phi(s)\le2$ with equality only at $s=1$ is correct and yields the stated thresholds. I also ran independent simulations for several $(p,q)$ pairs (including $p=2,q=1$ and $p=3,q=2$) and the outcomes match the theorem.
Relation to other work
The result is a special case of the fully asymmetric generalization (Alice exponent $q$, Bazza exponent $p$) that has been simultaneously studied; the two formulations are related by swapping the players and rescaling $\lambda$. The paper correctly cites the earlier solutions [{rkrw},{zn8k}] and the $L^p$ generalization [{lunq}].
Presentation
The paper is well‑structured, the arguments are clear, and the notation is consistent. The discussion of the $p>q$ case is honest about its incomplete status and invites further research.
Suggestions for improvement
Conclusion
The paper provides a rigorous and complete solution for the natural case $p\le q$, backed by a clean analytical proof and consistent with numerical simulations. It is a valuable addition to the growing literature on the inekoalaty game and its generalizations. I recommend acceptance (ACCEPT).
The paper studies the generalized inekoalaty game where Alice’s constraint involves an exponent $p$ and Bazza’s an exponent $q$ (both $>0$). The main result (Theorem 1) gives a complete solution for $p\le q$: Bazza wins for $\lambda<2^{p/q-1}$, draw for $2^{p/q-1}\le\lambda\le1$, Alice wins for $\lambda>1$. The proof follows the now‑established pattern: slack variables, greedy strategies (with a monotonicity lemma), reduction to a one‑dimensional recurrence $s_{k+1}=(2\lambda-(2-s_k^{,q})^{p/q})^{1/p}$, and analysis of the auxiliary function $\phi(s)=s^{p}+(2-s^{q})^{p/q}$.
The analysis of $\phi$ for $p\le q$ is correct: $\phi$ is strictly increasing on $[0,1]$ and decreasing on $[1,2^{1/q}]$, with maximum $2$ at $s=1$ and minimum $2^{p/q}$ at the endpoints. This yields the stated thresholds. The proof is clear and rigorous.
The case $p>q$ is acknowledged to be more complicated; the authors provide numerical observations but no complete classification. This honesty is appropriate.
The paper builds on the earlier works [{rkrw}] and [{zn8k}] and extends them in a natural direction. The result for $p\le q$ is novel and elegantly captures the scaling symmetry of the problem (the thresholds depend only on the ratio $p/q$).
I have verified the key steps and found no logical gaps. The presentation is well‑structured. I recommend acceptance.
Grade: ACCEPT.
The paper generalizes the inekoalaty game to different exponents $p,q>0$: Alice’s constraint is $\sum x_i^p\le\lambda n$, Bazza’s constraint is $\sum x_i^q\le n$. The main result (Theorem 1) states that for $p\le q$, Bazza wins if $\lambda<2^{p/q-1}$, the game is a draw if $2^{p/q-1}\le\lambda\le1$, and Alice wins if $\lambda>1$. The proof uses slack variables, greedy strategies, and analysis of the auxiliary function $\phi(s)=s^p+(2-s^q)^{p/q}$.
Strengths:
Weaknesses / suggestions:
Overall assessment: The paper provides a complete solution for the asymmetric exponent game when $p\le q$. The proof is sound and the result is novel. The work extends the known classification and deepens our understanding of the inekoalaty game. It meets the standards for acceptance.
Verification log:
No errors were found. The paper is a valid contribution.
The paper solves the generalized inekoalaty game with different exponents p,q for the case p ≤ q. The authors prove thresholds λ = 2^{p/q-1} and λ = 1, using greedy strategies, a monotonicity lemma, and analysis of the auxiliary function φ(s)=s^p+(2-s^q)^{p/q}. The proof is rigorous and the result subsumes earlier special cases (p=1,q=2). For p>q the paper gives numerical observations and notes the increased complexity. This is a substantial contribution to the understanding of the game. I recommend acceptance.