【24h】

How Many Potatoes Are in a Mesh?

机译:网格中有多少个土豆?

获取原文

摘要

We consider the combinatorial question of how many convex polygons can be made at most by using the edges taken from a fixed triangulation of n vertices. For general triangulations, there can be exponentially many: Ω( 1.5028~n) and O(1.62~n) in the worst case. If the triangulation is fat (every triangle has its angles lower-bounded by a constant δ> 0), then there can be only polynomially many: Ω(n~(1/2[2π/δ]) ) and O(n~([π/δ])). If we count convex polygons with the additional property that they contain no vertices of the triangulation in their interiors, we get the same exponential bounds in general triangulations, and Ω(n~([2π/3δ])) and O(n~([2π/3δ])) in fat triangulations.
机译:我们考虑一个组合问题,即通过使用从n个顶点的固定三角剖分中获得的边,最多可以制作多少个凸多边形。对于一般的三角剖分,可能有成倍的指数:在最坏的情况下,Ω(1.5028〜n)和O(1.62〜n)。如果三角剖分是胖的(每个三角形的角度都由一个恒定的δ> 0下限限定),则只能有多项式:Ω(n〜(1/2 [2π/δ]))和O(n〜 ([π/δ]))。如果我们计算凸多边形的额外属性是,它们的内部不包含三角剖分的顶点,则在一般三角剖分中,我们将得到相同的指数范围,以及Ω(n〜([2π/3δ]))和O(n〜( [2π/3δ]))在脂肪三角剖分中。

著录项

相似文献

  • 外文文献
  • 中文文献
  • 专利
获取原文

客服邮箱:kefu@zhangqiaokeyan.com

京公网安备:11010802029741号 ICP备案号:京ICP备15016152号-6 六维联合信息科技 (北京) 有限公司©版权所有
  • 客服微信

  • 服务号