二次Chvatal秩的0/1多面体
0/1 Polytopes with Quadratic Chvatal Rank
摘要：
对于一个多胞形P，Chvatal闭包P'是通过同时加强所有可行不等式cx <= b（含积分c）到cx <= floor（b）而得到的.直到达到P的积分壳所需的该过程的迭代次数被称为Chvatal秩。如果P是[0，1]^n的子集，那么已知O（n^2 log n）次迭代总是足够的（Eisenbrand and and Schulz（1999）），并且有时至少需要（1+1/e-o（1））n次迭代（Pokutta and Stauffer（2011）），在下限和上界之间留下巨大的差距。我们证明了存在一个包含在0/1立方体中的多面体，其Chvatal秩为Omega（n^2），将缺口缩小到对数因子。事实上，甚至一个超线性的下界被提到作为一个开放的问题，由几个作者。我们选择的P是半随机背包多面体和单个分数顶点的凸包。主要的技术成分是联系Chvatal秩同时丢番图逼近w.r.t.定义P的法向量的L1范数。
英文摘要：
For a polytope P, the Chvatal closure P' is obtained by simultaneously strengthening all feasible inequalities cx <= b (with integral c) to cx <= floor(b). The number of iterations of this procedure that are needed until the integral hull of P is reached is called the Chvatal rank. If P is a subset of [0,1]^n, then it is known that O(n^2 log n) iterations always suffice (Eisenbrand and Schulz (1999)) and at least (1+1/e-o(1))n iterations are sometimes needed (Pokutta and Stauffer (2011)), leaving a huge gap between lower and upper bounds. We prove that there is a polytope contained in the 0/1 cube that has Chvatal rank Omega(n^2), closing the gap up to a logarithmic factor. In fact, even a superlinear lower bound was mentioned as an open problem by several authors. Our choice of P is the convex hull of a semi-random Knapsack polytope and a single fractional vertex. The main technical ingredient is linking the Chvatal rank to simultaneous Diophantine approximations w.r.t. the L1-norm of the normal vector defining P.
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