登录
首页 » Fortran » Sparse-matrix-algorithm-

Sparse-matrix-algorithm-

于 2021-01-30 发布 文件大小:4948KB
0 283
下载积分: 1 下载次数: 54

代码说明:

  稀疏矩阵算法及其程序实现(杨绍祺+谈根林),有解决稀疏矩阵需要的朋友可以好好认真的看一下(Sparse matrix algorithm and program implementation (Yang Shaoqi+ Tan root forest), solve sparse matrix can be a good friend in need a serious look at)

文件列表:

稀疏矩阵算法及其程序实现(杨绍祺%2B谈根林).pdf,5484757,2010-10-05

下载说明:请别用迅雷下载,失败请重下,重下不扣分!

发表评论

0 个回复

  • Euler_DG_Quadrilateral_2D
    自己写了一个二维Euler方程的间断有限元程序 上次发了一个三角形单元的程序 因为不是曲边单元 所以在圆柱后面容易形成涡 现在把程序改为曲边四边形单元了 没有涡出现 单元是8节点四边形单元 节点编号顺序是 1 5 2 6 3 7 4 8 也就是四个角点依次 是1 2 3 4 然后是边的中点编号 5 6 7 8. 时间推进采用 Runge-Kutta 方法 数值通量采用全局Lax-Friedrichs通量 仍然不能捕捉激波 因为没有做重构或者加人工粘性 等这个做出来了 再发一次。 程序没有进行优化 比如说内存的消耗没有优化 比如直边单元的边界积分仍然采 用了曲边的积分方法 增加了计算量 比如面积分、线积分都是采用的是Gauss- Legendre-Lobatto积分 积分精度会比一般的Gauss-Legendre积分精度低一阶 等 等问题。 二维的 纯属交流性质 就没有考虑这些问题 ^_^ 如果物面全部是直边 那么只要改变一个参数N 就可以获得不同的计算精度 且具 有谱精度 因为单元的节点是Gauss-Legendre-Lobatto积分点。 其实就是谱元法 (物面是曲边的情况我不清楚是不是也可以通过提高基函数的阶数 也就是增加N 来提高计算精度)(Wrote a two-dimensional Euler equations with discontinuous finite element program Last made ​ ​ a triangular element of the program, not curved edge unit is so easy to form a vortex in the cylinder behind the Program to curved edge quadrilateral element vortices appear Unit is the order of 8-node quadrilateral element node number is 15,263,748 which is the four corners of the points in turn Is 1234 and then the side of the midpoint of the number 5678. Time promote the use of Runge-Kutta method Numerical flux of the overall situation of Lax-Friedrichs, flux Still can not capture the shock wave did not do the reconstruction or artificial viscosity do it Zaifayici. The program is not optimized for example, memory consumption is not optimized such as straight-edge boundary integral of the unit is still mining Integral method to increase the amount of computation such as surface integral with a curved edge, the line integral using the Gauss- The Legendr)
    2021-01-29 12:58:40下载
    积分:1
  • Monte-Carlo
    详细全面的蒙特卡洛PPT讲解以及MATLAB算例,帮助大家0基础快速掌握MC的精华所在。(Detailed and comprehensive Monte Carlo PPT to explain and MATLAB example, to help you basically grasp the essence of MC.)
    2021-03-09 15:09:27下载
    积分:1
  • uve_pls
    对一维光谱实现偏最小二乘回归进行UVE变量选择(UVE variable selection on the spectrum)
    2021-04-19 15:18:51下载
    积分:1
  • 大连理工大矩阵与值分析上机作业
    利用matlab软件进行数值编程,优化计算。(Matlab software is used for numerical programming and optimization calculation.)
    2020-06-22 04:40:02下载
    积分:1
  • calrescapind
    this source code is use to calculate resister capacitor and inductor
    2009-09-29 18:26:22下载
    积分:1
  • 7756
    感应双馈发电机系统的仿真,基于分段非线性权重值的Pso算法,计算多重分形非趋势波动分析matlab程序。( Simulation of doubly fed induction generator system, Based on piecewise nonlinear weight value Pso algorithm, Calculation multifractal detrended fluctuation analysis matlab program.)
    2017-05-12 16:30:34下载
    积分:1
  • yxyffbc(dsb)
    有限元方法fortran编程,包括二维三维线性非线性静态分析、稳定流瞬时流分析、振动问题、特征值问题、并行计算等。包含了所有子程序。(Finite element method fortran programming, including 2D and 3D linear and nonlinear static analysis, steady flow transient flow analysis, vibration problems, eigenvalue problems, parallel computing. Contains all subroutines.)
    2013-11-24 23:02:01下载
    积分:1
  • rotation
    fluent 旋转边界算例,可以了解运动机械在fluent模拟中的设置(fluent rotation the boundary example, you can understand the mechanical movement set in fluent simulation)
    2020-11-03 10:59:53下载
    积分:1
  • LS-SVMlab-Toolbox
    最小二乘支持向量机工具想的设计,及其使用工具箱的指导手册。(The least squares support vector machine tool design think its use toolbox guidance manual.)
    2013-01-16 20:36:33下载
    积分:1
  • Soct
    假设有一个能装入总体积为T的背包和n件体积分别为w1 , w2 , … , wn 的物品,能否从n件物品中挑选若干件恰好装满背包,即使w1 +w2 + … + wn=T,要求找出所有满足上述条件的解。例如:当T=10,各件物品的体积{1,8,4,3,5,2}时,可找到下列4组解:(1,4,3,2),(1,4,5),(8,2),(3,5,2)。 [需求分析](Suppose there are a load of the backpack of the total volume of T and n volumes were w1, w2, ... wn items, whether selected from n items a number of exactly filled backpack, even if w1+w2+ ...+ wn = T, find out the solution of all the above conditions are met. For example: when T = 10, the volume of items {1,8,4,3,5,2}, can be found in the following four solutions: (1,4,3,2), (1,4,5 ), (8,2), (3,5,2). [Needs Analysis])
    2012-06-14 16:25:10下载
    积分:1
  • 696516资源总数
  • 106914会员总数
  • 0今日下载