代码搜索：协方差矩阵
找到约 10,000 项符合「协方差矩阵」的源代码
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www.eeworm.com/read/254498/4382010
cs sample3_6.cs
/*
* 示例程序Sample3_6: Matrix类的托伯利兹矩阵求逆的埃兰特方法
*/
using System;
using CSharpAlgorithm.Algorithm;
namespace CSharpAlgorithm.Sample
{
class Class1
{
[STAThread]
static void Main(string
www.eeworm.com/read/475210/6794140
cpp aes.cpp
// AES.cpp : Defines the entry point for the console application.
#include "stdafx.h"
#include
#include
#define Nb 4 //状态矩阵的字数（列数）
#define Nk 4 //初始密钥的字数
#define Nr 10 /
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c gaussian.c
//Using Gaussian elimination method and Gaussian elimination method with row scaled method to solve the following tri-diagonal system
//
//for n=10 and 100.
//
//解决方案：由于使用高斯消去发的前提是，在运算过程中A矩阵的主对角
www.eeworm.com/read/395206/8190288
m cholesky.m
function L=Cholesky(A)
%对对称正定矩阵A进行Cholesky分解
n=length(A);
L=zeros(n);
for k=1:n
delta=A(k,k);
for j=1:k-1
delta=delta-L(k,j)^2;
end
if delta
www.eeworm.com/read/395080/8196980
java shortpath.java
/**
a //图的邻接矩阵
display //用来打印运算结果
path //存入v(i,j)和最短路径
v //记录最短路径的顶点信息
**/
public class Shortpath
{
public static int[] display = new int[10];
public static float[] p
www.eeworm.com/read/394381/8227553
m inv.m
%求逆矩阵
%用法 B=inv(A) 其中A为数值或符号方阵,B返回A的逆
%例如
% inv([1 2;3 4]) %数值
% syms a b c d;inv([[a,b;c,d]) %符号
%
%INV Matrix inverse.
% INV(X) is the inverse of the square
www.eeworm.com/read/394381/8227890
m rand.m
%R=rand(m,n) 生成(0,1)上均匀分布的m行n列随机矩阵
%RAND Uniformly distributed random numbers.
% RAND(N) is an N-by-N matrix with random entries, chosen from
% a uniform distribution on the interval (0.0,1.0
www.eeworm.com/read/193048/8256186
m cmpmatrix.m
function flag=cmpmatrix(a,b)
% cmpmatrix function
% 用于确定两个矩阵维数是否相等的比较函数
% Copyright 2000-2001 ECUST.
% $Revision: 1.1 $ $Date: 2002/02/11 21:58:12 $
if any(any(a~=b)),
flag='matrix i
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m qstable.m
function s=qstable(a)
% 确定系统矩阵a的稳定性
% Copyright 2000-2001 ECUST.
% $Revision: 1.1 $ $Date: 2001/12/12 14:11:48 $
system_root=eig(a),i=find(real(system_root)>0);
if ~isempty(i),s='system u
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txt 5.22.txt
Status AddTSM(TSMatrix &A,TSMatrix B)
/* 将三元组矩阵B加到A上: A=A+B */
{
int pa,pb,pc;
int i,j,x,sum;
if(A.mu!=B.mu||A.nu!=B.nu) return ERROR;
if(A.tu==0) {