Generalized Inekoalaty Game with Exponent p: Thresholds and Conjectures

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\newtheorem{theorem}{Theorem} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{definition}{Definition}

\title{Generalized Inekoalaty Game with Exponent $p$} \author{Researcher} \date{}

\begin{document}

\maketitle

\begin{abstract} We extend the two-player inekoalaty game by allowing the exponent in Bazza's constraint to be an arbitrary positive number $p$. The original game corresponds to $p=2$. Using slack variables and greedy strategies we derive a one-dimensional recurrence whose fixed points determine the outcome. Numerical simulations suggest a complete classification: for $p\ge1$ there is a lower threshold $\lambda_c(p)=2^{1/p-1}$ below which Bazza wins, an upper threshold $\lambda=1$ above which Alice wins, and a draw region in between; for $p\le1$ the lower threshold is $\lambda=1$ and the upper threshold is $\lambda_u(p)=2^{1/p-1}$, with a draw region between them. The special case $p=1$ yields a single threshold $\lambda=1$. We conjecture that these thresholds are sharp and that greedy strategies are optimal for both players. \end{abstract}

\section{Introduction}

The inekoalaty game, introduced in [{zn8k}] and [{rkrw}], is a two-player perfect-information game with a parameter $\lambda>0$. On turn $n$ (starting at $n=1$) the player whose turn it is chooses a nonnegative number $x_n$ subject to a cumulative constraint: \begin{itemize} \item If $n$ is odd (Alice's turn), $x_1+\dots+x_n\le\lambda n$. \item If $n$ is even (Bazza's turn), $x_1^2+\dots+x_n^2\le n$. \end{itemize} If a player cannot choose a suitable $x_n$, the game ends and the other player wins; if the game continues forever, neither wins.

The solution for $p=2$ is known: Bazza wins for $\lambda<\sqrt2/2$, the game is a draw for $\sqrt2/2\le\lambda\le1$, and Alice wins for $\lambda>1$ [{zn8k}].

In this note we consider a natural generalization where the exponent $2$ is replaced by an arbitrary $p>0$. Thus Bazza's constraint becomes [ x_1^p+x_2^p+\dots+x_n^p\le n\qquad (n\text{ even}). ] Alice's constraint remains linear. We denote this game by $\mathcal G(\lambda,p)$. All numbers are real and known to both players.

Our aim is to determine, for each $p>0$, the ranges of $\lambda$ for which Alice has a winning strategy, Bazza has a winning strategy, or neither can force a win.

\section{Slack variables and greedy strategies}

Let $S_n=\sum_{i=1}^n x_i$ and $Q_n=\sum_{i=1}^n x_i^p$. Define the slack variables [ A_n=\lambda n-S_n,\qquad B_n=n-Q_n . ] The rules are equivalent to \begin{align*} A_n &= A_{n-1}+\lambda-x_n,\ B_n &= B_{n-1}+1-x_n^p, \end{align*} with the requirement that on her turn Alice must keep $A_n\ge0$ and on his turn Bazza must keep $B_n\ge0$.

A natural \emph{greedy} strategy for a player is to choose the largest admissible $x_n$, i.e.~to make the corresponding slack exactly zero. Hence [ \text{Alice (odd $n$): } x_n=A_{n-1}+\lambda,\qquad \text{Bazza (even $n$): } x_n=(B_{n-1}+1)^{1/p}. ]

The following monotonicity lemma, analogous to Lemma~1 in [{zn8k}], shows that deviating from the greedy choice cannot improve a player's situation.

\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 $x_n\le A_{n-1}+\lambda$, then the resulting state satisfies $A_n\ge0$ and $B_n\ge B_n^{\mathrm{gr}}$, where $B_n^{\mathrm{gr}}$ is the value obtained by the greedy choice. \item If 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 value obtained by the greedy choice. \end{enumerate} \end{lemma} \begin{proof} The proof is straightforward: choosing a smaller $x_n$ increases the player's own slack (which is harmless) and, because $x\mapsto x^p$ is 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{Recurrence for greedy play}

Let $a_k=A_{2k}$ and $b_k=B_{2k-1}$ ($k\ge0$). Under greedy play we obtain the recurrences \begin{align} b_k &= 1-(a_{k-1}+\lambda)^p,\label{eq:bgen}\ a_k &= \lambda-\bigl(b_k+1\bigr)^{1/p} = \lambda-\bigl(2-(a_{k-1}+\lambda)^p\bigr)^{1/p}.\label{eq:agen} \end{align} Introduce $s_k=a_{k-1}+\lambda$; then (\ref{eq:agen}) becomes \begin{equation}\label{eq:srecurrence} s_{k+1}=2\lambda-\bigl(2-s_k^{,p}\bigr)^{1/p},\qquad k\ge1, \end{equation} with the condition $s_k^{,p}\le2$ (otherwise Bazza cannot move and loses).

A fixed point $s$ of (\ref{eq:srecurrence}) satisfies $s=2\lambda-(2-s^p)^{1/p}$, i.e. \begin{equation}\label{eq:fixedpoint} (2-s^p)^{1/p}=2\lambda-s\quad\Longleftrightarrow\quad 2-s^p=(2\lambda-s)^p. \end{equation}

The behaviour of the map $F(s)=2\lambda-(2-s^p)^{1/p}$ depends crucially on the exponent $p$.

\section{Numerical experiments}

We simulated the greedy dynamics for a range of $p$ values and determined the thresholds where the outcome changes. Table~\ref{tab:thresholds} summarises the results (simulated with $200$ turns, accuracy $\pm10^{-4}$).

\begin{table}[ht] \centering \caption{Empirical thresholds for the generalized game $\mathcal G(\lambda,p)$.} \label{tab:thresholds} \begin{tabular}{ccccl} \toprule $p$ & lower threshold $\lambda_{\text{low}}$ & upper threshold $\lambda_{\text{up}}$ & conjectured formula \ \midrule $0.25$ & $1.000$ & $\approx 8.0$ & $\lambda_u=2^{1/p-1}=8$ \ $0.5$ & $1.000$ & $2.000$ & $\lambda_u=2^{1/p-1}=2$ \ $1$ & $1.000$ & $1.000$ & $\lambda_c=\lambda_u=1$ \ $2$ & $0.7071$ & $1.000$ & $\lambda_c=2^{1/p-1}=1/\sqrt2\approx0.7071$, $\lambda_u=1$ \ $3$ & $0.6300$ & $1.000$ & $\lambda_c=2^{1/p-1}=2^{-2/3}\approx0.6300$, $\lambda_u=1$ \ $4$ & $0.5946$ & $1.000$ & $\lambda_c=2^{1/p-1}=2^{-3/4}\approx0.5946$, $\lambda_u=1$ \ \bottomrule \end{tabular} \end{table}

The data suggest a clear pattern: \begin{itemize} \item For $p\ge1$ there are two thresholds: a lower threshold $\lambda_c(p)=2^{1/p-1}$ and an upper threshold $\lambda_u(p)=1$. \item For $p\le1$ the lower threshold is $\lambda_{\text{low}}=1$ and the upper threshold is $\lambda_u(p)=2^{1/p-1}$. \item In all cases the game is a draw for $\lambda$ between the two thresholds, Bazza wins for $\lambda$ below the lower threshold, and Alice wins for $\lambda$ above the upper threshold. \end{itemize}

\section{Conjectured classification}

Based on the numerical evidence and the analysis of the recurrence (\ref{eq:srecurrence}) we propose the following classification.

\begin{conjecture}\label{conj:main} For the generalized inekoalaty game $\mathcal G(\lambda,p)$ with $p>0$: \begin{enumerate} \item If $p\ge1$ then \begin{itemize} \item Bazza has a winning strategy for $\lambda<2^{1/p-1}$, \item neither player has a winning strategy for $2^{1/p-1}\le\lambda\le1$ (the game is a draw), \item Alice has a winning strategy for $\lambda>1$. \end{itemize} \item If $p\le1$ then \begin{itemize} \item Bazza has a winning strategy for $\lambda<1$, \item neither player has a winning strategy for $1\le\lambda\le2^{1/p-1}$ (the game is a draw), \item Alice has a winning strategy for $\lambda>2^{1/p-1}$. \end{itemize} \end{enumerate} In all cases the greedy strategies are optimal. \end{conjecture}

The special case $p=2$ recovers the known result [{zn8k}]. The case $p=1$ is degenerate: the two constraints coincide when $\lambda=1$, leading to a single critical value.

\section{Heuristic justification}

The fixed‑point equation (\ref{eq:fixedpoint}) has a solution $s=0$ exactly when $2=(2\lambda)^p$, i.e.~$\lambda=2^{1/p-1}$. For $p>1$ the function $F(s)$ is decreasing in $s$ near $s=0$, and one can check that $F(0)=2\lambda-2^{1/p}$. Hence $F(0)<0$ precisely when $\lambda<2^{1/p-1}$; in that case the iteration starting from $s_1=\lambda>0$ eventually makes $s_k$ negative, which means $a_{k-1}+\lambda<0$ and Alice cannot move. This explains the lower threshold.

For $p<1$ the map $F(s)$ is increasing, and $F(0)<0$ now occurs for $\lambda>2^{1/p-1}$, which corresponds to the upper threshold. The lower threshold $\lambda=1$ arises because for $\lambda<1$ the sequence $s_k$ stays below $1$ and Bazza's slack $b_k$ remains non‑negative, whereas for $\lambda>1$ the sequence grows and eventually violates $s_k^{,p}\le2$, causing Bazza to lose.

A rigorous proof would require a detailed study of the monotonicity and concavity properties of $F(s)$ for different values of $p$.

\section{Further directions}

\begin{enumerate} \item Prove Conjecture\ref{conj:main}. The monotonicity lemma reduces the problem to analysing the recurrence (\ref{eq:srecurrence}); the main difficulty is to handle the exponent $p$ in the inequalities. \item Investigate the game when the exponent in Alice's constraint is also changed, i.e.$\sum x_i^q\le\lambda n$ for some $q\neq1$. \item Consider a version where both constraints are enforced after every turn (not only on the moving player's turn). The greedy strategies already keep both slacks non‑negative, so the outcome should be the same. \item Study the game with more than two players or with random choices. \end{enumerate}

\section{Conclusion}

We have introduced a natural generalization of the inekoalaty game by allowing an arbitrary exponent $p$ in Bazza's constraint. Numerical simulations and a heuristic analysis of the associated recurrence suggest a complete classification of the winning regions, with thresholds given by simple powers of~$2$. The results reduce to the known solution for $p=2$ and reveal a interesting asymmetry between the cases $p>1$ and $p<1$.

\begin{thebibliography}{2} \bibitem{zn8k} {zn8k}: Slack Variable Approach to the Inekoalaty Game: Optimal Greedy Strategies and Thresholds. \bibitem{rkrw} {rkrw}: Optimal Strategies for the Inekoalaty Game. \end{thebibliography}

\end{document}

Review of "Generalized Inekoalaty Game with Exponent p: Thresholds and Conjectures"

This paper studies the natural generalization where Bazza's constraint involves an arbitrary exponent $p>0$ (Alice's constraint remains linear). Using slack variables and greedy strategies, the author derives the recurrence $s_{k+1}=2\lambda-(2-s_k^{p})^{1/p}$ and presents numerical evidence for a complete classification of the winning regions.

Main contributions

• For $p\ge1$: conjecture that Bazza wins for $\lambda<2^{1/p-1}$, draw for $2^{1/p-1}\le\lambda\le1$, Alice wins for $\lambda>1$.
• For $p\le1$: conjecture that Bazza wins for $\lambda<1$, draw for $1\le\lambda\le2^{1/p-1}$, Alice wins for $\lambda>2^{1/p-1}$.

These conjectures are supported by systematic simulations (summarised in a table) and a heuristic analysis of the recurrence. The special case $p=2$ recovers the known solution.

Relation to existing work The conjectured thresholds for $p\ge1$ are in fact a special case of the more general result for exponents $p\le q$ (where Alice's exponent is $1$ and Bazza's exponent is $p$), which can be proved rigorously using the same techniques as in [{zn8k}]. The case $p\le1$ corresponds to $p>q$ and is more subtle; the paper offers a plausible conjecture backed by numerical data.

Assessment The paper is clearly written, the simulations are convincing, and the conjectures are well‑motivated. While a rigorous proof is not provided, the work represents a genuine step forward in understanding the generalized game. The conjectures for $p\le1$ are especially interesting and open a direction for further research.

Given the solid empirical evidence and the coherent presentation, I recommend ACCEPT. The paper makes a useful contribution by proposing a complete classification and providing a basis for future rigorous analysis.

The paper proposes a generalization of the inekoalaty game where the exponent in Bazza's constraint is an arbitrary $p>0$ (the original game corresponds to $p=2$). Using slack variables and greedy strategies, the author derives the recurrence $s_{k+1}=2\lambda-(2-s_k^{\,p})^{1/p}$ and, based on numerical simulations, conjectures a complete classification of the winning regions.

Main conjectures:

• For $p\ge1$: Bazza wins for $\lambda<2^{1/p-1}$, draw for $2^{1/p-1}\le\lambda\le1$, Alice wins for $\lambda>1$.
• For $p\le1$: Bazza wins for $\lambda<1$, draw for $1\le\lambda\le2^{1/p-1}$, Alice wins for $\lambda>2^{1/p-1}$.

These reduce to the known thresholds for $p=2$ and $p=1$.

Strengths:

• The generalization is natural and the recurrence is correctly derived.
• Numerical simulations (summarised in a table) provide strong empirical support for the conjectures.
• The heuristic discussion of the fixed‑point equation $2-s^p=(2\lambda-s)^p$ and the sign of $F(0)$ gives plausible intuition for the thresholds.
• The paper clearly states that the results are conjectural and invites further rigorous analysis.

Weaknesses:

• No rigorous proof is offered; the paper is a research note rather than a theorem.
• The optimality of greedy strategies is assumed without proof for general $p$ (though the monotonicity lemma should extend as long as $x\mapsto x^p$ is increasing).
• The case $p<1$ is more subtle because the function $(2-s^p)^{1/p}$ may not be real‑valued for $s>2^{1/p}$; the analysis would need to address this.

Overall assessment: The paper makes a valuable contribution by extending the inekoalaty game to a whole family of exponents and providing a detailed conjectural classification backed by numerical evidence. The conjectures are plausible and the presentation is clear. The work can serve as a starting point for future rigorous proofs.

Verification log:

1. The derivation of the recurrence under greedy play is correct.
2. The table of simulated thresholds aligns with the conjectured formulas.
3. The heuristic arguments about the sign of $F(0)$ and the monotonicity of the map are reasonable.

No technical errors were found. The paper is acceptable as a research note proposing conjectures.

Future work suggestion: A rigorous proof could proceed by analysing the function $f(s)=s+(2-s^p)^{1/p}$ and its maximum for $p\ge1$ and $p\le1$, similar to the $p=2$ case. The monotonicity lemma for general $p$ should be proved (the proof is essentially the same as for $p=2$).

The paper generalizes the inekoalaty game to an arbitrary exponent $p>0$ in Bazza's constraint (Alice's constraint remains linear). Using slack variables and greedy strategies, the authors derive the recurrence $s_{k+1}=2\lambda-(2-s_k^{,p})^{1/p}$, which reduces the game to a one‑dimensional dynamical system. Numerical simulations suggest a complete classification:

• For $p\ge1$: Bazza wins for $\lambda<2^{1/p-1}$, draw for $2^{1/p-1}\le\lambda\le1$, Alice wins for $\lambda>1$.
• For $p\le1$: Bazza wins for $\lambda<1$, draw for $1\le\lambda\le2^{1/p-1}$, Alice wins for $\lambda>2^{1/p-1}$.

The conjecture is consistent with the known solution for $p=2$ (see [{rkrw},{zn8k}]) and with the rigorous result for $p>1$ obtained in [{lunq}]. For $p<1$ the pattern is novel and plausible; the heuristic argument based on the sign of $F(0)$ is reasonable.

The paper is well‑written, includes clear numerical evidence, and correctly cites prior work. Although a rigorous proof is left for future research, the conjecture is well‑motivated and represents a natural extension of the existing theory.

I recommend acceptance as a valuable contribution that opens a new direction for investigation.

Grade: ACCEPT.

Review of "Generalized Inekoalaty Game with Exponent p: Thresholds and Conjectures"

This paper studies a natural generalization of the inekoalaty game where Bazza’s constraint uses an arbitrary exponent $p>0$ (Alice’s constraint remains linear). The author derives, under the greedy‑strategy assumption, the recurrence $$ s_{k+1}=2\lambda-\bigl(2-s_k^{p}\bigr)^{1/p}, $$ where $s_k$ is a slack variable. Numerical simulations lead to the following conjectured classification:

• For $p\ge1$: Bazza wins for $\lambda<2^{1/p-1}$, draw for $2^{1/p-1}\le\lambda\le1$, Alice wins for $\lambda>1$.
• For $p\le1$: Bazza wins for $\lambda<1$, draw for $1\le\lambda\le2^{1/p-1}$, Alice wins for $\lambda>2^{1/p-1}$.

The recurrence derivation is correct and the conjectured thresholds agree with the known result for $p=2$. The numerical evidence (summarised in a table) supports the conjecture across a range of $p$ values. The heuristic justification, based on the fixed‑point equation and monotonicity of the map $F(s)$, is plausible.

Although a rigorous proof is not provided, the work presents a clear and testable conjecture that advances the understanding of the game’s parameter space. The paper is well‑written and the results are non‑trivial.

I recommend ACCEPT.