凸体的覆盖与填装

发布时间：2018-07-01 14:32

本文选题：覆盖 + 填装　；参考：《河北师范大学》2016年硕士论文

【摘要】：设C,D是平面凸多边形,G1,G2,…是C的位似拷贝.若D(?)UGNn,则称{Cn}覆盖D.若D(?)UCn且{Ci}两两内部不交,则称{Cn}可填装到D.特别地,当C有-条边与D的一条边平行时,称{Cn}平行覆盖或填装D.论文第二章主要考虑用等边三角形覆盖与填装单位正方形,并得到以下两个结论：任意(有限或无限)等边三角形序列,若它的面积之和不小于2+(?)3,则它可平行覆盖单位正方形；任意(有限或无限)等边三角形序列,若它的面积之和不超过(?)/6,则它可平行填装到单位正方形.在论文的第三章考虑了用正方形序列覆盖上底为1,2,高为(?)/2的等腰梯形,且得到以下两个结论：任意(有限或无限)正方形序列,若它的面积之和不小于4,则它可平行覆盖直角边为1,(?)的直角三角形；任意(有限或无限)正方形序列,若它的面积之和不小于4,则它可平行覆盖上底为1,2,高为(?)/2的等腰梯形.设是单位超立方体H的一个覆盖.若不存在超立方体集S'能覆盖H,其中且s(C''i)s(Ci)(其中s(Ci)表示超立方体Ci的边长),则称S是H的最小覆盖.令是n个超立方体对单位超立方体的最小覆盖}.第四章考虑在d-维欧几里得空间Ed中的d维超立方体的覆盖问题(其中d≥4).即用较小的df维超立方体覆盖d维单位超立方体.并得到以下结论：若C是d维单位超立方体的最小覆盖且C有n个d-维超立方体,则gd(n)≤s(C)；当n≥2d+1,有gd(n)≤2d-11+δ,其中δ是趋于0的正实数；对任意n≥2d,有gd(n)≥2d-1.
[Abstract]:Let's set C, D is a plane convex polygon, G1, G2,... It is a similar copy of C. If D (?) UGNn, then called {Cn} to cover D. if D (?) UCn and {Ci} 22 do not cross, then the {Cn} can be filled to D. especially, when the C is parallel to the one side of the D, it is called parallel coverage or packing paper second chapters mainly to cover and fill the square with the equilateral triangle, and get the following two conclusions: An equilateral triangular sequence of meaning (finite or infinite), if its area is not less than 2+ (?) 3, it can cover a unit square in parallel; an arbitrary (finite or infinite) equilateral triangle sequence, if its area is not more than (?) /6, can be filled in parallel to a single square. In the third chapter of the paper, the upper bottom is covered with a square sequence. For 1,2, a (?) /2 isosceles trapezoid, and the following two conclusions are obtained: an arbitrary (finite or infinite) square sequence, and if its area is not less than 4, it can cover a right angle triangle with a right angle of 1 and (?); an arbitrary (finite or infinite) square sequence, if its area is not less than 4, it can cover a parallel bottom to 1,2, The isosceles trapezoid of high (?) /2 is a cover of unit hypercube H. If no hypercube set S'can cover H, and S (C''i) s (Ci) (s (Ci) represents the length of the hypercube Ci), it is called the minimum coverage of the hypercube. The coverage problem of D dimensional hypercubes in Ed (d > 4). That is to use smaller DF dimensional hypercubes to cover D dimensional hypercubes. And the following conclusions are obtained: if C is the minimum coverage of D dimensional hypercubes and C has n d- dimensional hypercubes, Gd (n) is less than equal. For any n more than 2D, there is Gd (n) or more 2d-1.
【学位授予单位】：河北师范大学
【学位级别】：硕士
【学位授予年份】：2016
【分类号】：O186.5

【相似文献】