本文選題：覆蓋 + 填裝　； 參考：《河北師范大學(xué)》2016年碩士論文

【摘要】：設(shè)C,D是平面凸多邊形,G1,G2,…是C的位似拷貝.若D(?)UGNn,則稱{Cn}覆蓋D.若D(?)UCn且{Ci}兩兩內(nèi)部不交,則稱{Cn}可填裝到D.特別地,當(dāng)C有-條邊與D的一條邊平行時(shí),稱{Cn}平行覆蓋或填裝D.論文第二章主要考慮用等邊三角形覆蓋與填裝單位正方形,并得到以下兩個(gè)結(jié)論：任意(有限或無(wú)限)等邊三角形序列,若它的面積之和不小于2+(?)3,則它可平行覆蓋單位正方形；任意(有限或無(wú)限)等邊三角形序列,若它的面積之和不超過(guò)(?)/6,則它可平行填裝到單位正方形.在論文的第三章考慮了用正方形序列覆蓋上底為1,2,高為(?)/2的等腰梯形,且得到以下兩個(gè)結(jié)論：任意(有限或無(wú)限)正方形序列,若它的面積之和不小于4,則它可平行覆蓋直角邊為1,(?)的直角三角形；任意(有限或無(wú)限)正方形序列,若它的面積之和不小于4,則它可平行覆蓋上底為1,2,高為(?)/2的等腰梯形.設(shè)是單位超立方體H的一個(gè)覆蓋.若不存在超立方體集S'能覆蓋H,其中且s(C''i)s(Ci)(其中s(Ci)表示超立方體Ci的邊長(zhǎng)),則稱S是H的最小覆蓋.令是n個(gè)超立方體對(duì)單位超立方體的最小覆蓋}.第四章考慮在d-維歐幾里得空間Ed中的d維超立方體的覆蓋問(wèn)題(其中d≥4).即用較小的df維超立方體覆蓋d維單位超立方體.并得到以下結(jié)論：若C是d維單位超立方體的最小覆蓋且C有n個(gè)d-維超立方體,則gd(n)≤s(C)；當(dāng)n≥2d+1,有g(shù)d(n)≤2d-11+δ,其中δ是趨于0的正實(shí)數(shù)；對(duì)任意n≥2d,有g(shù)d(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.
【學(xué)位授予單位】：河北師范大學(xué)
【學(xué)位級(jí)別】：碩士
【學(xué)位授予年份】：2016
【分類號(hào)】：O186.5

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