1) Ding Fang
丁方
1.
Ding Fang, as one of the most extraordinary artists, reflects 1980s China culture trend, even more he can be recognized as a symbol of Chinese contemporary art explorer of the time.
本文把丁方置于1980年代的中国现代艺术背景下,通过对他的艺术母题、艺术风格和艺术追求进行剖析,寻找他的艺术与当时社会的文艺思潮之间的关系。
3) Latin square
拉丁方
1.
Analysis of balance coding of Latin squares;
拉丁方的平衡编码特性分析
2.
Searching Standard Latin Square Algorithm with Order;
利用位序法求标准拉丁方算法
4) Latin squares
拉丁方
1.
In this paper we give a new definition of uniform cyclic Latin squares and denote by UCL(n) simply,and give UCL(n) of order n≤10 by the minimal fluctuation criteria.
给出了循环均匀拉丁方的新定义，简记为ＵＣＬ（ｎ），并给出了ｎ≤１０的循环均匀拉丁方。
2.
An efficient algorithm for finding the isotopy class of the latin squares is given.
提出了一种拉丁方合痕分类的快速算法 ,该算法结构简单 ,复杂度低 。
3.
A P~2-order (P is odd prime) orthogonal Latin squares was employed as a constructor for building three different types of magic squares:quadratically and doubly-magic squares,and magic cubes.
讨论用同一种模式的 P2 (P为奇素数 )阶正交拉丁方为数理座标 ,构造 P2 阶二次、双重和立体三种不同幻方的方法 。
5) strong latin square
强拉丁方
1.
The concepts of strong latin square and strong latin moment are introduced in this paper.
本文引入了强拉丁方和强拉丁矩的概念 ,证明了当 m≥ 2且为偶数时 ,强拉丁矩的数目是 (2 m) !·2 m ( m -1 ) / 2 ,如果我们不考虑同构 ,有 (2 m ) !· 2 m ( m -1 ) / 2 -(m-1)· 2 m -1 ) ( m -2 ) / 2 个竞赛图 ,且完全图 k2 m +1 有 2 m ( m -1 ) / 2个相互不同构的竞赛
2.
In order to solve this problem, this paper introduces the concepts of strong Latin square and strong Latin moment, and we change alspach′s enumerating problem into the enumerating problems of strong Latin square and strong Latin moment.
为了解决这个问题 ,本文引入了强拉丁方和强拉丁矩的概念 ,我们把 Alspach的计数问题变成了强拉丁方和强拉丁矩的计数问题。
6) Latin cube
拉丁立方
1.
A proof of a method of constructing mutually orthogonal Latin cube of odd order;
奇N阶三重正交拉丁立方构造方法的论证
2.
Latin cube and relative definitions by extending the corresponding ones of Latin square are introduced.
对拉丁方的概念进行推广，把构成拉丁方的二维矩阵提升到三维，引入了拉丁立方及相关概念；并给出了进行ｎ 阶规范拉丁立方的计数与合痕分类的算法；利用此算法，得出１ 至５ 阶规范拉丁立方的数目及其合痕类类
3.
Searching the method of constructing triple mutually orthogonal Latin cubes of odd order is carried out by the author of this paper.
介绍了奇 n阶三重正交拉丁立方的构造方法 ,并利用所述的方法构造出 1 3阶三重正交拉丁立方。
补充资料：丁方
1.四方。
说明：补充资料仅用于学习参考，请勿用于其它任何用途。
参考词条