Resolving the Schwartz Quadratic Meander Number Conjecture
解决Schwartz二次曲折数猜想
Charles Daly, Diaaeldin Taha
AI总结 通过定义循环排列的曲折数,证明其最大值在n的二次函数范围内,解决了Schwartz关于拓扑推销员问题的猜想,并构造了从线性到二次增长率的连续族。
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一个循环曲折是平面上嵌入的有向环,它与一条固定无限直线或圆横截相交于$2n$个线性有序点。通过记录环访问这些点的顺序,循环曲折在这些标记点上诱导出一个循环排列。相应地,给定一个$n$个字母上的排列,可以问是否存在一个循环曲折以这种方式诱导该排列,如果不存在,允许更多交点时最有效的方式是什么?这个过程为$n$个字母上的排列赋予了一个复杂度度量。本文的主要结果表明,所有$n$个字母上的循环排列的这个量(称为曲折数)的最大值在$n$的二次函数范围内有上下界。这一结果解决了Schwartz~\cite{richtpss}关于拓扑推销员问题的猜想。最后,我们构造了$n$个字母上的循环排列族,其曲折数实现了从线性到二次的增长率的连续谱。
A cyclic meander is an embedded oriented loop in the plane intersecting a fixed infinite line, or circle, transversely in a linearly ordered set of $2n$ points. By keeping track of the order in which the loop visits these points, the cyclic meander induces a cyclic permutation on these marked points. Correspondingly, given a permutation on $n$ letters, one can ask whether or not a cyclic meander induces the permutation in this manner, and if not, what is the most efficient way of doing so if we allow more points of intersection? This process gives a way of associating to a permutation on $n$ letters a measurement of complexity of the permutation in question. The principal result of this work shows that the maximum of this quantity, the \emph{meander number}, over all cyclic permutations on $n$ letters, is bounded above and below quadratically in $n$. This result resolves a conjecture of Schwartz~\cite{richtpss} in relation to his work on the topological salesman problem. We conclude this work by constructing families of cyclic permutations on $n$ letters whose meander numbers realize a continuum of growth rates between linear and quadratic.