← 研踪 PaperTrailarXiv:2611.06003 · v1
How Unbalanced Must a Latin Rectangle Be?
首次提交 2026-11-04 13:52
版本更新 2026-11-04 13:52
雷达观测 2026-11-05 11:10
摘要
We view a $k\times n$ Latin rectangle as $n$ points in the grid $[n]^k$. Each symbol gives one point, whose coordinates are the columns it occupies. The rectangle is balanced when these points are pairwise equidistant in the $\ell_1$ metric. Its imbalance measures the distance from this ideal. We study the minimum imbalance from three sides. Constructively, we use a simulated annealing that moves symbols within rows and only penalises column clashes. Against the reference table of Voboril et al. (2025), it improves 33 of the 51 non-square values for $8\le n\le 13$, by up to 39.4%, and matches the rest. A cheap invariant, the row group, certifies that none of the 33 new rectangles is isotopic to rows of a group table, so the earlier column-swapping local search cannot reach any of them. Exactly, a computer-assisted branch and bound proves that the minimum imbalance of a $3\times 9$ rectangle is $56$. Díaz et al. found this value, but its optimality was open. A sharper pruning bound, derived from the identity $I=2\sum(d-t)^+$, reduces the search tree by a factor of $2.4$ to $3.3$ on the instances with $k\ge3$ that we solve. Asymptotically, the minimum imbalance is $Θ(n^3)$ for every fixed $k$. A pigeonhole argument gives a lower constant of order $k\,4^{-k}$, and random rectangles give an upper constant of order $\sqrt k$. For $k=2$, the limit of $\min I/n^3$ exists, a permuton attains it, and it lies between $4/375$ and $0.1196$. Perfect balance is an $\ell_1$-equilateral problem, so it forces $k=Ω(n/\log n)$ rows. For $2\times n$ rectangles, single two-minute annealing runs improve five of the seven six-hour MIP upper bounds for $14\le n\le 20$. Finally, the Lean 4 kernel checks every rectangle reported in this paper against a 74-line specification that imports only Lean's core library. We also prove the pigeonhole lower bound in Lean with Mathlib.
观测到的版本
相邻记录可能跳过中间版本；首次发现 v2 及以上也会标为版本更新。
v1 · 2026-11-04 13:52 · 最新观测版本
How Unbalanced Must a Latin Rectangle Be?
We view a $k\times n$ Latin rectangle as $n$ points in the grid $[n]^k$. Each symbol gives one point, whose coordinates are the columns it occupies. The rectangle is balanced when these points are pairwise equidistant in the $\ell_1$ metric. Its imbalance measures the distance from this ideal. We study the minimum imbalance from three sides. Constructively, we use a simulated annealing that moves symbols within rows and only penalises column clashes. Against the reference table of Voboril et al. (2025), it improves 33 of the 51 non-square values for $8\le n\le 13$, by up to 39.4%, and matches the rest. A cheap invariant, the row group, certifies that none of the 33 new rectangles is isotopic to rows of a group table, so the earlier column-swapping local search cannot reach any of them. Exactly, a computer-assisted branch and bound proves that the minimum imbalance of a $3\times 9$ rectangle is $56$. Díaz et al. found this value, but its optimality was open. A sharper pruning bound, derived from the identity $I=2\sum(d-t)^+$, reduces the search tree by a factor of $2.4$ to $3.3$ on the instances with $k\ge3$ that we solve. Asymptotically, the minimum imbalance is $Θ(n^3)$ for every fixed $k$. A pigeonhole argument gives a lower constant of order $k\,4^{-k}$, and random rectangles give an upper constant of order $\sqrt k$. For $k=2$, the limit of $\min I/n^3$ exists, a permuton attains it, and it lies between $4/375$ and $0.1196$. Perfect balance is an $\ell_1$-equilateral problem, so it forces $k=Ω(n/\log n)$ rows. For $2\times n$ rectangles, single two-minute annealing runs improve five of the seven six-hour MIP upper bounds for $14\le n\le 20$. Finally, the Lean 4 kernel checks every rectangle reported in this paper against a 74-line specification that imports only Lean's core library. We also prove the pigeonhole lower bound in Lean with Mathlib.
查看此版本解读 →查看此版本原文 ↗