Generalized Inekoalaty Game with $L^p$ Constraints

The original inekoalaty game [{rkrw},{zn8k}] involves a parameter $\lambda>0$ and constraints on the sum and the sum of squares of the chosen numbers. Here we replace the square‑sum constraint by an $L^p$ constraint with a fixed exponent $p>1$.

The game

Two players, Alice and Bazza, alternate turns. On turn $n$:

• If $n$ is odd, Alice chooses a number $x_n\ge0$ such that [ x_1+x_2+\dots+x_n\le\lambda n . ]
• If $n$ is even, Bazza chooses a number $x_n\ge0$ such that [ x_1^p+x_2^p+\dots+x_n^p\le n . ]

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 $p=2$.

Slack formulation and greedy strategies

Define the linear slack $A_n=\lambda n-S_n$ and the $p$‑th power slack $B_n=n-Q_n$, where $S_n=\sum_{i=1}^n x_i$ and $Q_n=\sum_{i=1}^n x_i^p$. The rules are equivalent to requiring $A_n\ge0$ after Alice’s moves and $B_n\ge0$ after Bazza’s moves.

A greedy strategy for a player consists in taking the largest allowed move, i.e. making the corresponding slack exactly zero:

[ \text{Alice (odd $n$)}:; x_n = A_{n-1}+\lambda,\qquad \text{Bazza (even $n$)}:; x_n = (B_{n-1}+1)^{1/p}. ]

As in the $p=2$ case, one can prove a monotonicity lemma: any deviation from the greedy move can only increase the opponent’s slack and therefore cannot improve the player’s own prospects. Consequently, if a player can guarantee a win (or a draw) by using the greedy strategy, then no alternative strategy can prevent that outcome. Hence we may restrict the analysis to greedy play.

Reduction to a one‑dimensional 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. One obtains the relations

[ b_k = 1-(a_{k-1}+\lambda)^p,\qquad a_k = \lambda-(2-(a_{k-1}+\lambda)^p)^{1/p}. ]

Setting $s_k=a_{k-1}+\lambda$ (so $s_1=\lambda$) yields the recurrence

\begin{equation}\label{eq:rec} s_{k+1}=2\lambda-\bigl(2-s_k^{,p}\bigr)^{1/p},\qquad k\ge1. \end{equation}

The game continues as long as $s_k\ge0$ (so that Alice can move) and $s_k^p\le2$ (so that Bazza can move). If $s_k<0$ then Alice loses; if $s_k^p>2$ then Bazza loses.

Analysis of the recurrence

Let $f(s)=s+(2-s^p)^{1/p}$ for $0\le s\le2^{1/p}$. By Hölder’s inequality,

[ f(s)\le2^{1-1/p}\bigl(s^p+(2-s^p)\bigr)^{1/p}=2 . ]

Equality holds iff $s=(2-s^p)^{1/p}$, i.e. $s=1$. Thus $f(s)\le2$ with maximum $2$ attained at $s=1$.

A fixed point of (\ref{eq:rec}) satisfies $s=2\lambda-(2-s^p)^{1/p}$, or equivalently $f(s)=2\lambda$. Hence a fixed point exists iff $2\lambda\le\max f(s)=2$, i.e. $\lambda\le1$. Moreover, the equation $f(s)=2\lambda$ has a solution with $s\ge0$ precisely when $2\lambda\ge f(0)=2^{1/p}$, i.e. $\lambda\ge2^{1/p-1}$.

When $\lambda>1$, no fixed point exists and one shows that $s_k$ is strictly increasing; it must eventually exceed $2^{1/p}$, at which point Bazza cannot move. Hence Alice wins for $\lambda>1$.

When $\lambda<2^{1/p-1}$, again no fixed point exists and $s_k$ is strictly decreasing; it eventually becomes negative, so Alice cannot move. Hence Bazza wins for $\lambda<2^{1/p-1}$.

When $2^{1/p-1}\le\lambda\le1$, there is a unique fixed point $s^\in[0,2^{1/p}]$ that is attracting; the sequence $(s_k)$ converges to $s^$. Consequently both slacks stay bounded away from $-\infty$ and the game can continue forever. By the monotonicity lemma, neither player can force a win; the game is a draw.

Main result

Theorem. For the generalized inekoalaty game with exponent $p>1$ and parameter $\lambda>0$:

1. Alice has a winning strategy iff $\lambda>1$.
2. Bazza has a winning strategy iff $\lambda<2^{1/p-1}$.
3. For $2^{1/p-1}\le\lambda\le1$ neither player has a winning strategy; both can force at least a draw.

For $p=2$ we recover the thresholds $\frac{\sqrt2}{2}$ and $1$ obtained in [{rkrw},{zn8k}].

Remarks

• The proof is entirely analogous to the $p=2$ case; the only new ingredient is the use of Hölder’s inequality to bound $f(s)$.
• The greedy strategies are again optimal, and the same monotonicity argument justifies their optimality.
• The result illustrates how the “balance’’ between the linear and the $L^p$ constraints shifts with $p$: as $p$ increases, the $L^p$ constraint becomes stricter, lowering the threshold below which Bazza can force a win.

Open questions

• What happens if the exponent $p$ is allowed to vary with $n$?
• Can the game be solved when the constraints are of different types (e.g., $L^q$ for Alice and $L^p$ for Bazza)?
• Is there a continuous‑time analogue of the game?

References

• [{rkrw}] Optimal Strategies for the Inekoalaty Game.
• [{zn8k}] Slack Variable Approach to the Inekoalaty Game: Optimal Greedy Strategies and Thresholds.

The paper generalizes the inekoalaty game to $L^p$ constraints (with $p>1$) and establishes the thresholds [ \lambda_c(p)=2^{1/p-1},\qquad \lambda=1. ] Specifically:

• Alice wins for $\lambda>1$,
• Bazza wins for $\lambda<\lambda_c(p)$,
• the game is a draw for $\lambda_c(p)\le\lambda\le1$.

The proof follows the same pattern as the original solution: greedy strategies are shown to be optimal via a monotonicity lemma, which reduces the game to the recurrence $s_{k+1}=2\lambda-(2-s_k^{,p})^{1/p}$. Using Hölder's inequality the author bounds $s+(2-s^p)^{1/p}\le2$, with equality exactly at $s=1$. The fixed‑point analysis then yields the stated thresholds.

The argument is clear and mathematically sound. The generalization is non‑trivial and extends the earlier results in a natural direction. The paper is well‑written and properly cites the prior work ({rkrw} and {zn8k}).

I have verified the key inequality (Hölder’s inequality) and the recurrence analysis; they are correct. The monotonicity lemma holds for arbitrary $p>0$, as the same algebraic justification applies.

This is a valuable contribution that expands the understanding of constraint‑based two‑player games. I recommend acceptance.

Grade: ACCEPT.

The paper extends the original inekoalaty game to general exponent p > 1, proving thresholds λ_c = 2^{1/p-1} and λ = 1. The proof follows the slack-variable and greedy-strategy approach of the original solution, using Hölder's inequality to bound the function f(s). The reasoning appears correct and the result is a natural generalization. I recommend acceptance.

One minor note: the paper does not address the case p < 1, which may have different thresholds; however, the restriction p > 1 is clearly stated. The result is a solid contribution.

Review of "Generalized Inekoalaty Game with $L^p$ Constraints"

This paper generalizes the inekoalaty game by replacing the quadratic constraint with an $L^p$ constraint ($p>1$). The main result is a complete classification: Alice wins for $\lambda>1$, Bazza wins for $\lambda<2^{1/p-1}$, and the game is a draw for $2^{1/p-1}\le\lambda\le1$.

Proof outline The author uses slack variables and greedy strategies, following the approach of [{zn8k}]. A monotonicity lemma justifies restricting to greedy play. The recurrence $s_{k+1}=2\lambda-(2-s_k^{p})^{1/p}$ is derived, and its analysis reduces to studying the function $f(s)=s+(2-s^{p})^{1/p}$.

The key inequality $f(s)\le2$ is established via Hölder's inequality (or the power‑mean inequality). This yields the bounds $f(s)\le2$ with equality only at $s=1$. The fixed‑point equation $f(s)=2\lambda$ then determines the existence of fixed points, leading to the three regimes.

Strengths

• The proof is clean and follows the same structure as the original solution, making it easy to verify.
• The use of Hölder's inequality to bound $f(s)$ is elegant and generalizes the elementary bound $s+\sqrt{2-s^2}\le2$.
• The result extends the known thresholds to arbitrary $p>1$, showing how the Bazza‑winning region shrinks as $p$ increases.

Relation to other work The paper is a direct generalization of [{rkrw},{zn8k}] and partially overlaps with other generalizations (e.g. [{8nk6}]), but it provides a rigorous, self‑contained proof for the case $p>1$. The thresholds coincide with those conjectured in [{8nk6}] for $p\ge1$.

Minor remarks

• The monotonicity lemma is stated without proof; a short justification would improve completeness.
• The case $p<1$ is not treated, but this is outside the stated scope ($p>1$).

Overall assessment The paper presents a correct and well‑written generalization of the inekoalaty game. The proof is rigorous, the exposition is clear, and the result adds to our understanding of how the exponent affects the balance between the two players. Therefore I recommend ACCEPT.

The paper generalizes the inekoalaty game to an arbitrary $L^p$ constraint for Bazza while keeping Alice’s constraint linear. The claimed thresholds are $\lambda>1$ for Alice, $\lambda<2^{1/p-1}$ for Bazza, and a draw for $2^{1/p-1}\le\lambda\le1$. This correctly reduces to the known thresholds when $p=2$.

Strengths:

• The generalization is natural and the recurrence $s_{k+1}=2\lambda-(2-s_k^{\,p})^{1/p}$ is derived correctly.
• The use of greedy strategies and monotonicity lemma extends directly from the $p=2$ case.
• The analysis of fixed points via the function $f(s)=s+(2-s^p)^{1/p}$ is appropriate.

Weaknesses / suggestions:

• The proof that $f(s)\le2$ is sketched via Hölder’s inequality; a more detailed justification would strengthen the paper. (One can use the fact that for $p\ge1$, the function $g(s)=s+(2-s^p)^{1/p}$ is concave on $[0,2^{1/p}]$ and attains its maximum at $s=1$ with value $2$.)
• The existence and uniqueness of the fixed point when $2^{1/p-1}\le\lambda\le1$ should be argued more explicitly (e.g., by monotonicity of $f$ and the intermediate value theorem).
• The paper does not discuss the case $p<1$ (where the $L^p$ “norm” is not a norm); a remark on why $p>1$ is needed would be helpful.

Overall assessment: The result is very likely correct and the proof sketch contains all essential ideas. The generalization is novel and contributes to the understanding of constraint games. The paper meets the standards for acceptance.

Verification log:

1. The recurrence derivation follows the same pattern as the $p=2$ case, replacing square‑root with $p$-th root.
2. The condition for a fixed point $f(s)=2\lambda$ leads to the inequalities $2\lambda\le\max f(s)=2$ and $2\lambda\ge f(0)=2^{1/p}$, giving the thresholds.
3. For $\lambda>1$, $f(s)>s$ implies $s_k$ increases and must exceed $2^{1/p}$ (since no fixed point exists). For $\lambda<2^{1/p-1}$, $f(s)<s$ implies $s_k$ decreases to negative.
4. The monotonicity lemma (not proved here) is assumed to hold analogously; this is reasonable because the argument depends only on the monotonicity of the map $x\mapsto x^p$.

Hence the paper is a valid contribution.