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fecgm
- 独立成份分析（ＩＣＡ）以及ｗｉｎｎｅｒ滤波 Source separation of complex signals with JADE. Jade performs `Source Separation in the following sense: X is an n x T data matrix assumed modelled as X = A S + N where o A is an unknown n x m matrix with full rank.
fisher
- 费希尔线性判别分析代码 Find the Fisher linear separator w (a column vector). X is is the training set (X is a matrix. Each row of X is a vector containing the features of a single sample). y is a column vector with the labels of the training set (1
siyuanshu2
- 不同坐标系下相同测量点之间的四元数转换矩阵求解，可以此来求解坐标系A到坐标系B之间的转换矩阵-Different coordinates between the same measurement points quaternion transformation matrix solution can be used to solve this coordinate system A to coordinate system transformation matrix between B
GaussElimination
- 使用高斯消元法，解线性方程组。写作AX=B型的矩阵形式解决。-Using Gaussian elimination, solution of linear equations. Writing AX = B type of matrix solution.
Gauss_Elimination
- 使用高斯消元法，解线性方程组。写作AX=B型的矩阵形式解决。-Using Gaussian elimination, solution of linear equations. Writing AX = B type of matrix solution.
A_LU
- bool lu(double *a, int *pivot, int n)；矩阵的LU分解。 假设数组an*n在内存中按行优先次序存放，此函数使用高斯列选主元消去法，将其就地进行LU分解。pivot为输出函数.pivot[0,n)中存放主元的位置排列. 函数成功时返回false，否则返回true. bool guass(double const *lu, int const *p, double *b, int n) 求线性方程组的解。 假设矩阵lum*n为某个矩阵a
A_QR
- void qr(double *a, double *d, int n) 矩阵的QR分解 假设数组an*n在内存中按行优先次序存放，此函数使用HouseHolder变换将其就地进行QR分解。 d为输出参数，d[0,n)存放QR分解的上三角矩阵对角线元素。 bool householder(double const *qr, double const *d, double *b, int n) 求线性代数方程组的解。 假设矩阵qrn*n为某个矩阵an*n的QR分解，在内
shortpath
- 甲城市与乙城市之间共有n 座城市，互相以公路连通。甲城市、乙 城市以及其它各城市之间的公路连通情况及每段公路的长度由矩阵 M1 给出。 每段公路均由地方政府收取不同额度的养路费等费用，具体数额由矩 阵M2 给出。 请给出在需付养路费总额不超过1500 的情况下，该公司货车运送其 产品从甲城市到乙城市的最短运送路线。-A city with b city between each other with city, n. road. A city, b City and
final.asm
- a)实现时钟功能，可以在两个七段数码管上显示秒钟时间或者分钟时间，用一个开关控制两者的切换。 b)实现闹钟功能，时间到播放一段音乐，并在发光二极管上播放走马灯图案，在双色点阵发光二极管上滚动显示自己的学号。能控制滚动显示的速度以及音乐播放的速度，且用一个开关控制闹钟的开关。-a) achieve clock function, can be in two seven-segment digital tube display seconds or minutes, with a switch
maxsize
- 稀疏矩阵采用三元组表示。 （1）求两个具有相同行列数的稀疏矩阵A和B的相加矩阵C，并输出C。 （2）求出C的转置矩阵D，输出D。-Sparse matrix expressed by triples. (1) Find the ranks of the two have the same number of sparse matrix A and B of the sum matrix C, and the output C. (2) find the C of the transp
Cellular-Neural-Network
- 细胞神经网络(CNN)是一种和人类神经网络非常相似的并行计算模型，各个邻接节点间有不同的通信。在本程序中A模型是反馈矩阵，B是控制矩阵。-Cellular neural network (CNN) is very similar to the human neural network model of parallel computation, all adjacent nodes have different communication. A model of this process is
MatrixSerialMultiply1
- c++程序分别从文件1和文件2中读取矩阵A和B，进行相乘之后将矩阵C写到文件3中.-c++ program reads matrix a and b from file1 and file2 distributedly,then run matrix multiply and get matrix c,write the matrix c into file3.
pls
- 输入自变量与因变量，输出x。y主成分、负荷及回归系数- Inputs: x x matrix y y matrix Outputs: t score for x p loading for x u score for y q loading for y b regression coefficient
pls_copy
- 这是一个非线性回归偏最小二乘程序，输入因变量与自变量，输出为x，y的主成分与负荷因子与回归系数- Inputs: x x matrix y y matrix Outputs: t score for x p loading for x u score for y q loading for y b regression coefficient
originalsimpleM111.m
- 原始单纯形法（大M法，无需给出初始基变量）。输入：C是n维行向量，A是m*n的系数矩阵，b是m维列向量 输出：x最优解（如果有的话），fval最优值，flag解的状态说明，interation求解时的循环次数-The original simplex method (big M method, without giving the initial basic variable). Input: C is the n-dimensional row vector, A is the coef
1
- 非编码键盘是利用MCS—51单片机内部的定时/计数器、中断系统、以及外围的按键和LED数码管显示等部件，设计一个单片机非编码矩阵键盘。并能通过按键实现显示0、1、2、3、4、5、6、7、8、9、A、B、C、D、E、F。-The keyboard is used non-coding MCS-51 microcontroller internal timer/counter, interrupt system and the external buttons and LED digital dis
caculator
- 利用12个按键（4*3矩阵键盘0~b，因为仅有这点按键）和一块4位七段数码管（用于显示） 能实现实数的加减乘除运算，精确至小数点后四位，并完整地显示输入及结果。 -Using the 12 keys (4* 3 matrix keypad 0 ~ b, because only this key) and a 4 segment digital tube (for display) to achieve real number addition and subtraction multi
cubesum
- Suppose there is a X x Y x Z 3D matrix A of numbers having coordinates (i, j, k) where 0 ≤ i < X, 0 ≤ j < Y, 0 ≤ k < Z. Now another X x Y x Z matrix B is defined from A such that the (i, j, k) element of B is the sum of all the the numbers i
lisanshuxushiyan3
- 以偶对的形式输入一个无向简单图的边，建立该图的邻接矩阵，判断图是否连通（A）。并计算任意两个结点间的距离（B）。对不连通的图输出其各个连通支（C）。-Even on the form to input an undirected graph edge, the establishment of the adjacency matrix to determine whether the connectivity graph (A). And calculate any distance betwe
Microsoft
- 设计一个矩阵相乘的程序，首先从键盘输入两个矩阵a，b的内容，并输出两个矩阵，输出ab－1结果。-Design a matrix multiplication of the program first and foremost from a keyboard two matrix a, b, and output two matrix and output result. ab － 1