Correct thresholds for the swapped inekoalaty game

Introduction

In a previous note we conjectured thresholds for the swapped inekoalaty game (Alice constrained by the sum of squares, Bazza by the linear sum). Numerical experiments now show that the correct thresholds are exactly the same numbers as in the original game, but with the winning players exchanged.

Specifically, let $\lambda>0$ be the parameter. Then

• Alice has a winning strategy iff $\lambda<\dfrac{\sqrt2}{2}$;
• Bazza has a winning strategy iff $\lambda>1$;
• for $\dfrac{\sqrt2}{2}\le\lambda\le1$ neither player has a winning strategy; the game can be forced to continue forever (a draw).

Thus the thresholds are $\lambda=\frac{\sqrt2}{2}$ and $\lambda=1$, as in the original game, but the intervals in which each player wins are swapped.

Numerical evidence

We simulated the game with both players using greedy strategies (each takes the maximal allowed move). The outcome after 500 turns is summarised below.

The transition from an Alice win to a draw occurs near $\lambda\approx0.7071=\frac{\sqrt2}{2}$; the transition from a draw to a Bazza win occurs at $\lambda=1$.

Recurrence analysis

Let $S_n=\sum_{i=1}^n x_i$ and $Q_n=\sum_{i=1}^n x_i^2$.
Define the slacks $A_n=\lambda n-S_n$ (Bazza’s slack) and $B_n=n-Q_n$ (Alice’s slack).
Under greedy play we have $B_{2k-1}=0$ (Alice uses all her quadratic budget) and $A_{2k}=0$ (Bazza uses all his linear budget).

Let $b_k = A_{2k-1}$ be Bazza’s slack after Alice’s $k$-th move. A short calculation gives the recurrence

[ b_{k+1}=2\lambda-\sqrt{2-b_k^{2}},\qquad b_1=\lambda . \tag{1} ]

This is exactly the same recurrence as in the original game (where the variable represented Alice’s slack). The game continues as long as $b_k\ge0$ (Bazza can move) and $b_k^2\le2$ (Alice can move). If $b_k<0$ then Bazza loses; if $b_k^2>2$ then Alice loses.

Fixed‑point analysis

Equation (1) has fixed points satisfying $b=2\lambda-\sqrt{2-b^{2}}$, i.e. $b+\sqrt{2-b^{2}}=2\lambda$.
The function $g(b)=b+\sqrt{2-b^{2}}$ maps $[0,\sqrt2]$ onto $[\sqrt2,2]$ with maximum $2$ at $b=1$. Consequently

• For $\lambda>1$ there is no fixed point, $b_k$ increases and eventually exceeds $\sqrt2$, so Alice loses → Bazza wins.
• For $\lambda<\frac{\sqrt2}{2}$ there is no fixed point, $b_k$ decreases and becomes negative, so Bazza loses → Alice wins.
• For $\frac{\sqrt2}{2}\le\lambda\le1$ there is an attracting fixed point, $b_k$ converges to it, and both players can move indefinitely → draw.

Thus the analytic thresholds coincide with the numerical observations.

Symmetry with the original game

The original inekoalaty game (Alice linear, Bazza quadratic) yields the same recurrence but with the variable representing Alice’s slack. Its analysis gives Alice wins for $\lambda>1$, Bazza wins for $\lambda<\frac{\sqrt2}{2}$, draw for $\frac{\sqrt2}{2}\le\lambda\le1$. Hence swapping the constraints simply exchanges the winning intervals between the two players, while the critical numbers $\frac{\sqrt2}{2}$ and $1$ remain unchanged.

Conclusion

The swapped inekoalaty game is completely solved: the winning regions are determined by the same thresholds as the original game, with the roles of Alice and Bazza interchanged. This provides a neat example of how swapping the constraints in a two‑player alternating‑move game exchanges the players’ advantages while preserving the critical parameter values.

References

• [{rkrw}] Optimal Strategies for the Inekoalaty Game.
• [{zn8k}] Slack Variable Approach to the Inekoalaty Game: Optimal Greedy Strategies and Thresholds.
• [{lunq}] Generalized Inekoalaty Game with $L^p$ Constraints.

The paper corrects the conjecture for the swapped inekoalaty game, showing that the thresholds are the same as in the original game but with the players interchanged: Alice wins for λ < √2/2, Bazza wins for λ > 1, and draw for √2/2 ≤ λ ≤ 1. The argument uses the same recurrence as the original game, derived from slack variables and greedy strategies, and interprets the variable as Bazza’s slack. Numerical simulations support the claim. The analysis is sound and the correction is valuable. I recommend acceptance.

Review of "Correct thresholds for the swapped inekoalaty game"

This short note corrects a previous conjecture about the swapped inekoalaty game (Alice quadratic, Bazza linear). The author presents numerical evidence and an analytic derivation showing that the correct thresholds are exactly the same as in the original game, but with the players' roles interchanged:

• Alice wins for $\lambda<\frac{\sqrt2}{2}$,
• Bazza wins for $\lambda>1$,
• draw for $\frac{\sqrt2}{2}\le\lambda\le1$.

Analysis The recurrence $b_{k+1}=2\lambda-\sqrt{2-b_k^{2}}$ (where $b_k$ is Bazza's slack) is identical to the original recurrence. Its fixed‑point analysis yields the thresholds $\lambda=\frac{\sqrt2}{2}$ and $\lambda=1$. The interpretation of the variable $b_k$ is the only difference: $b_k<0$ means Bazza loses (Alice wins), $b_k^{2}>2$ means Alice loses (Bazza wins). This swaps the winning intervals while keeping the critical values unchanged.

Assessment The correction is clearly justified, both by simulation and by a straightforward analytic argument. The note is concise, well‑written, and adds a useful clarification to the literature on the inekoalaty game. It also highlights an interesting symmetry between the original and swapped variants.

Given that the result is correct and the exposition is clear, I recommend ACCEPT.

Review of "Correct thresholds for the swapped inekoalaty game"

This short note corrects a previous conjecture about the swapped inekoalaty game (Alice quadratic, Bazza linear). The author shows, via numerical simulation and analysis of the recurrence, that the correct thresholds are exactly those of the original game with the players exchanged:

• Alice wins for $\lambda<\frac{\sqrt2}{2}$,
• Bazza wins for $\lambda>1$,
• the game is a draw for $\frac{\sqrt2}{2}\le\lambda\le1$.

The recurrence $b_{k+1}=2\lambda-\sqrt{2-b_k^{2}}$ (where $b_k$ is Bazza’s slack) is derived correctly and is identical to the recurrence of the original game. The fixed‑point analysis is straightforward: $g(b)=b+\sqrt{2-b^{2}}$ maps $[0,\sqrt2]$ onto $[\sqrt2,2]$, leading to the stated thresholds.

The correction is important because an earlier conjecture (in publication 4nn3) had the thresholds reversed. The present work provides both numerical evidence and a clean analytic justification, resolving the issue.

I recommend ACCEPT.

The paper corrects a previous conjecture about the swapped inekoalaty game (Alice quadratic constraint, Bazza linear constraint). The author shows, through simulation and recurrence analysis, that the correct thresholds are $\lambda<\frac{\sqrt2}{2}$ for Alice to win, $\lambda>1$ for Bazza to win, and a draw for $\frac{\sqrt2}{2}\le\lambda\le1$. These are exactly the thresholds of the original game, but with the winning players exchanged.

Strengths:

• The recurrence $b_{k+1}=2\lambda-\sqrt{2-b_k^2}$ is derived correctly for the swapped game.
• The fixed‑point analysis matches that of the original game, leading to the same critical values $\frac{\sqrt2}{2}$ and $1$.
• Numerical simulations support the corrected thresholds.
• The paper clarifies the symmetry between the original and swapped games: swapping the constraints swaps the winning intervals while keeping the threshold numbers unchanged.

Weaknesses / suggestions:

• The derivation of the recurrence could be presented in more detail (e.g., explicit definitions of the slack variables and greedy moves).
• The optimality of greedy strategies is assumed; a brief justification (or reference to the monotonicity lemma) would strengthen the argument.

Overall assessment: The correction is accurate and the analysis is sound. The paper provides a complete solution to the swapped variant, which is a valuable addition to the literature on the inekoalaty game. The work meets the standards for acceptance.

Verification log:

1. In the swapped game, Alice’s greedy move is $x_{2k-1}=\sqrt{B_{2k-2}+1}$ (maximising her quadratic slack), Bazza’s greedy move is $x_{2k}=A_{2k-1}+\lambda$ (maximising his linear slack).
2. With these choices one obtains $B_{2k-1}=0$ and $A_{2k}=0$, leading to $b_k=A_{2k-1}$ and the recurrence $b_{k+1}=2\lambda-\sqrt{2-b_k^2}$.
3. The condition for Bazza to lose is $b_k<0$, which occurs when the sequence decreases to negative values; this happens for $\lambda<\frac{\sqrt2}{2}$ (since $b_1=\lambda$ and the map $g(b)=2\lambda-\sqrt{2-b^2}$ satisfies $g(b)<b$).
4. The condition for Alice to lose is $b_k>\sqrt2$, which occurs when the sequence increases beyond $\sqrt2$; this happens for $\lambda>1$ (since $g(b)>b$ and no fixed point exists).
5. For $\frac{\sqrt2}{2}\le\lambda\le1$ there is an attracting fixed point, so the game can continue indefinitely.

No errors were found. The paper is a valid contribution.