大小为2的匹配的精确有序Ruzsa-Szemerédi数
Ordered Ruzsa-Szemeredi Numbers at Matching Size Two
- Wuhan University（武汉大学）
- The University of Manchester（曼彻斯特大学）
机构由 AI 辅助整理，请以论文原文为准。
AI总结：
该研究确定了匹配大小为2时有序Ruzsa-Szemerédi数在n=5至19的精确值，缩小了n=20的范围，给出渐近构造，揭示了序11是孤立例外，序20的值仍未解决。
AI中文摘要：
有序Ruzsa-Szemerédi图是一种图，其边集被划分为等大小的匹配，每个匹配由以其起始的排序后缀诱导产生。Behnezhad和Ghafari引入这类图来参数化完全动态匹配的更新时间，但人们对该数本身几乎一无所知。记f(n)为匹配大小为2时的最大部分数，我们确定了n从5到19的每个序对应的f(n)精确值，将序20的范围缩小到两个连续值，并给出了显式渐近构造。核心是有序分解与完全图的K₄剥离之间的双射，每一步删除当前构成团的四个顶点中的完美匹配。这立刻得到计数下界floor(n(n-4)/4)，并将等式问题简化为是否可达三次或近三次余数。结构引理将候选缩小到无桥连通图，收缩对应关系将奇序映射到大1的偶序，剩下的有限情形分析通过无同构的反向搜索完成。该下界仅在序5至9和11时达到，其余所有达到的序均恰好差1，因此序11是孤立例外而非奇偶现象：自然的等式猜想不成立且呈不规则失效。上界通过对完全三次计数的闭失败扫描验证，每个分解均由独立验证器对照定义重新检查，序20的两个值究竟取哪个仍未解决。
英文摘要：
Bondy and Szwarcfiter defined $\mathrm{ex}^*(n,F)$ as the largest number of edges in an $n$-vertex graph whose edge set partitions into induced copies of $F$; for $F=2K_2$ the deficiency $\binom{n}{2}-\mathrm{ex}^*(n,2K_2)$ is $Θ(n^{3/2})$. We study the ordered relaxation at fixed matching size, in which each part need only be induced in the union of itself with the parts that follow it; write $\mathrm{ORS}_n(r)$ for the largest number of parts, so that $r\,\mathrm{ORS}_n(r)$ is the ordered analogue of $\mathrm{ex}^*(n,rK_2)$. Our main tool is a characterisation valid for every $r$: an ordered decomposition into induced $r$-matchings is a sequence of steps that start from $K_n$ and repeatedly delete a perfect matching from $2r$ vertices currently spanning a clique. Reading a decomposition backwards turns a condition about the ordering into a reachability question that an exhaustive search can settle. For $r=2$ we determine $\mathrm{ORS}_n(2)$ exactly at orders five through nineteen, where it takes the values $1,3,5,8,11,14,19,23,28,34,40,47,54,62,70$, and we confine $\mathrm{ORS}_{20}(2)$ to $\{78,79\}$. The counting bound $\lfloor n(n-4)/4\rfloor$ is attained at orders five through nine and at eleven, and missed by exactly one part at every other order below twenty, so order eleven is an isolated exception, not a parity effect. Across this range the ordered deficiency equals $\frac32n+O(1)$, and along powers of two a dyadic construction keeps it below $O(n\log n)$; whether it is linear for all $n$ is our main open question. The structural results are formalised in Lean 4, and the searches are certified by fail-closed sweeps and an independent checker.
