solver_mod

一维热传导方程的求解，包括显式和隐式，及其对比(One-dimensional heat conduction equations, including explicit and implicit, and the contrast)

流体力学中有限体积法的求解程序，matlab代码-Finite volume method for solving procedures, matlab code(matlab code, -Finite, volume, method, for, solving, procedures, MATLAB, code)

主要功能是进行高斯正算，高斯反算及透视方位投影的显示。(Main function is to be considered are Gaussian, inverse Gaussian and the perspective projection display position.)

说明：  利用GA遗传算法解决欺骗函数最优问题，具体问题描述如下，如有问题请与我联系(The deceptive functions are a family of functions in which there exists low-order building blocks that do not combine to form the higher-order building blocks. Here, a deceptive problem that consists of 25 copies of the order-4 fully deceptive function DF2 is constructed for this paper. DF2 can be described as follows: f(0000)=28 f(0001)=26 f(0010)=24 f(0011)=18 f(0100)=22 f(0101)=6 f(0110)=14 f(0111)=0 f(1000)=20 f(1001)=12 f(1010)=10 f(1011)=2 f(1100)=8 f(1101)=4 f(1110)=6 f(1111)=30 This problem has a maximal function value of 750.)

洛斯阿拉莫斯实验室开发的显示动力有限元程序dyna3d，可处理碰撞，侵彻和接触等问题，有丰富的材料模型和状态方程，以及可扩展的二次开发能力。(source code 1 file)

电磁场圣经，林为干院士的书,学习电磁场重要材料(EMF Bible, it' s dry academician of the book, an important material for studying the electromagnetic field)

动力缩聚后的结构刚度、质量矩阵，以及频率(Dynamic stiffness after condensation, mass matrix, and the frequency)

说明：  用于abaqus二次开发材料VUMAT，为初级学者找到更简单的方法(Materials for ABAQUS secondary development)

说明：  Dijkstra算法与Floyd算法的matlab实现，以求解最短路问题(Dijkstra algorithm and Floyd algorithm matlab realize in order to solve the shortest path problem)

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])