#include "stdafx.h"
#include <math.h>
// 矩陣特徵值
int Maximum(double a[], int n)
{
int p=0;
int i;
for (i=1;i<n;i++) {
if (abs(a[i])>abs(a[p]))
p = i;
}
return p;
}
void MiChengMethod()
{
printf("求矩陣特徵值: 冪乘法\n");
double A[3][3] = {{4,-1,0},{0,4,-1},{0,-1,4}};
double x0[3] = {0,0,1};
double x1[3];
double lamda0,lamda1;
int k,i,j;
int n=3;
double epsilon = 1e-3;
int p;
p = Maximum(x0,n);
lamda0 = x0[p];
for (i=0;i<n;i++)
x0[i] /= lamda0;
k = 0;
for (;;) {
for (i=0;i<n;i++) {
x1[i] = 0;
for (j=0;j<n;j++)
x1[i] += A[i][j]*x0[j];
}
p = Maximum(x1,n);
lamda1 = x1[p];
k++;
if (k>1&&abs(lamda1-lamda0)<epsilon)
{
for (i=0;i<n;i++) x1[i]=x1[i]/lamda1; // 特徵向量
break;
}
else
{
lamda0 = lamda1; // 迭代
for (i=0;i<n;i++) x0[i]=x1[i]/lamda1;
}
}
PS(&k,1,"迭代次數");
PS(&lamda1,1,"按模特徵值lamda");
PS(x1,n,"特徵向量");
}
void PingYiYuanDianMethod()
{
printf("冪法的加速: 平移原點法\n");
double A[3][3] = {{4,-1,0},{0,4,-1},{0,-1,4}};
double x0[3] = {0,0,1};
double x1[3];
double lamda0,lamda1;
int k,i,j;
int n=3;
double epsilon = 1e-3;
int p;
double mu=2.9; // 平移量
for (i=0;i<n;i++) // 平移
A[i][i]-=mu;
p = Maximum(x0,n);
lamda0 = x0[p];
for (i=0;i<n;i++)
x0[i] /= lamda0;
k = 0;
for (;;) {
for (i=0;i<n;i++) {
x1[i] = 0;
for (j=0;j<n;j++)
x1[i] += A[i][j]*x0[j];
}
p = Maximum(x1,n);
lamda1 = x1[p];
k++;
if (k>1&&abs(lamda1-lamda0)<epsilon)
{
for (i=0;i<n;i++) x1[i]=x1[i]/lamda1; // 特徵向量
lamda1 += mu;
break;
}
else
{
lamda0 = lamda1; // 迭代
for (i=0;i<n;i++) x0[i]=x1[i]/lamda1;
}
}
PS(&k,1,"迭代次數");
PS(&lamda1,1,"按模特徵值lamda");
PS(x1,n,"特徵向量");
}
void AitkenJiaSuMethod()
{
printf("冪法的加速: 埃特肯加速法\n");
double A[3][3] = {{4,-1,0},{0,4,-1},{0,-1,4}};
double x0[3] = {0,0,1};
double x1[3];
double lamda[3];
int k,i,j;
int n=3;
double epsilon = 1e-3;
int p;
double lam[2]={0}; // 加速
double maxv;
p = Maximum(x0,n);
lamda[0] = x0[p];
for (i=0;i<n;i++)
x0[i] /= lamda[0];
k = 0;
for (;;)
{
for (i=0;i<n;i++) {
x1[i] = 0;
for (j=0;j<n;j++)
x1[i] += A[i][j]*x0[j];
}
p = Maximum(x1,n);
maxv = x1[p];
k++;
if (k<3)
lamda[k] = maxv;
else
{
lamda[0]=lamda[1];
lamda[1]=lamda[2];
lamda[2]=maxv;
}
if (k>2) // 加速公式
{
lam[0] = lam[1];
lam[1] = lamda[0]
-(lamda[1]-lamda[0])*(lamda[1]-lamda[0])/(lamda[2]-2*lamda[1]+lamda[0]);
}
if (k>3&&abs(lam[1]-lam[0])<epsilon)
break;
else
for (i=0;i<n;i++) x0[i]=x1[i]/maxv;
}
PS(&k,1,"迭代次數");
printf("按模特徵值lamda=%.7f\n",lam[1]);
}
void FanMiMethod()
{
printf("反冪法: 求矩陣最小特徵值\n");
double a[3][3] = {{1,-1,2},{-2,0,5},{6,-3,6}};
double b[3] = {1,1,1};
double l[3][3],u[3][3];
double y[3]={0};
int n=3;
int i,j;
double x[3];
double lamda0,lamda1;
double epsilon=1e-3;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
l[i][j] = u[i][j] = 0.;
}
}
// LU分解
int k = 0;;
int r;
for (k = 0; k < n; k++) {
// 先計算u
for (j = k; j < n; j++) {
u[k][j] = a[k][j];
for (r = 0; r < k && k > 0; r++) {
u[k][j] -= l[k][r]*u[r][j];
}
}
// 後計算l
l[k][k] = 1.;
for (i = k+1; i < n; i++) {
l[i][k] = a[i][k];
for (r = 0; r < k && k > 0; r++) {
l[i][k] -= l[i][r]*u[r][k];
}
l[i][k] /= u[k][k];
}
}
printf("l=\n");
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
printf("%f\t",l[i][j]);
printf("\n");
}
printf("u=\n");
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
printf("%f\t",u[i][j]);
printf("\n");
}
int p= Maximum(b,n);
lamda0=b[p];
for (i=0;i<n;i++) {
b[i]/=lamda0;
}
// 迭代
k=0;
for(;;)
{
// 回代求解
// 回代 求y
// Ly = b
y[0] = b[0];
for (i = 1; i < n; i++) {
y[i] = b[i];
for (j = 0; j < i; j++) {
y[i] = y[i] - l[i][j]*y[j];
}
}
// 求x
//Ux=y
x[n-1] = y[n-1]/u[n-1][n-1];
for (i = n-2; i >=0; i--) {
x[i] = y[i];
for (j = i+1; j < n; j++) {
x[i] = x[i] - u[i][j]*x[j];
}
x[i] /= u[i][i];
}
k++;
p=Maximum(x,n);
lamda1 =x[p];
printf("t%d=%f\n",k,lamda1);
printf("y%d= ",k);
for (i = 0; i < n; i++) {
printf("%.4f\t",x[i]/lamda1);
}
printf("\n");
if (abs(lamda1-lamda0)<epsilon)
{
break;
}
else
{
lamda0=lamda1;
for (i=0;i<n;i++) {
b[i]=x[i]/lamda1;
}
}
}
}
void FanMiMethod_2()
{
printf("反冪法: 已知某特徵值的近似值求更精確的特徵值\n");
double a[3][3] = {{4,-1,0},{0,4,-1},{0,-1,4}};
double b[3] = {0,0,1};
double l[3][3],u[3][3];
double y[3]={0};
int n=3;
int i,j;
double x[3];
double lamda0,lamda1;
double epsilon=1e-3;
double lam=4.93;
for (i = 0; i < n; i++) {
a[i][i] -= lam; // 減掉對角元
}
// LU分解, 先算u,後算l
int k = 0;;
int r;
for (k = 0; k < n; k++) {
// 先計算u, 行k
for (j = 0; j < n; j++) {
if (j<k)
u[k][j]=0.;
else
{
u[k][j] = a[k][j];
for (r = 0; r < k && k > 0; r++) {
u[k][j] -= l[k][r]*u[r][j];
}
}
}
// 後計算l,列k
for (i = 0; i < n; i++)
{
if (i<k) l[i][k]=0.;
else if(i==k) l[i][k]=1.;
else
{
l[i][k] = a[i][k];
for (r = 0; r < k && k > 0; r++) {
l[i][k] -= l[i][r]*u[r][k];
}
l[i][k] /= u[k][k];
}
}
}
printf("l=\n");
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
printf("%f\t",l[i][j]);
printf("\n");
}
printf("u=\n");
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
printf("%f\t",u[i][j]);
printf("\n");
}
int p= Maximum(b,n);
lamda0=b[p];
for (i=0;i<n;i++) {
b[i]/=lamda0;
}
// 迭代
k=0;
for(;;)
{
// 回代求解
// 回代 求y
// Ly = b
y[0] = b[0];
for (i = 1; i < n; i++) {
y[i] = b[i];
for (j = 0; j < i; j++) {
y[i] = y[i] - l[i][j]*y[j];
}
}
// 求x
//Ux=y
x[n-1] = y[n-1]/u[n-1][n-1];
for (i = n-2; i >=0; i--) {
x[i] = y[i];
for (j = i+1; j < n; j++) {
x[i] = x[i] - u[i][j]*x[j];
}
x[i] /= u[i][i];
}
k++;
p=Maximum(x,n);
lamda1 =x[p];
printf("t%d=%f\n",k,lamda1);
// printf("t%d=%f\n",k,1./lamda1+lam);
printf("y%d= ",k);
for (i = 0; i < n; i++) {
printf("%f\t",x[i]/lamda1);
}
printf("\n");
if (abs(lamda1-lamda0)<epsilon)
break;
else
{
lamda0=lamda1;
for (i=0;i<n;i++) {
b[i]=x[i]/lamda1;
}
}
}
printf("更精確的特徵值=%f\n",1./lamda1+lam);
}
// 將矩陣化成二次型的方法
void JacobiMethod()
{
printf("求實對稱矩陣特徵值的雅可比方法:\n");
double a[3][3] = {{-1,2,1},{2,-4,1},{1,1,-6}};
double b[3][3];
int i,j;
int p,q;
int n=3;
double epsilon = 5e-10;
double w;
double mu,lamda,s,c;
int k=0;
double B[3][3]={{1,0,0},{0,1,0},{0,0,1}}; // 特徵向量
double U[3][3]={{1,0,0},{0,1,0},{0,0,1}}; // 旋轉矩陣
double Tmp[3][3];
printf("輸入矩陣a%d=\n",k);
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
printf("%f\t",a[i][j]);
printf("\n");
}
for (;;) {
// 找到矩陣中非對角元最大值位置(p,q)
p = 0;
q = 1;
for (i=0;i<n;i++) {
for (j=0;j<n;j++) {
if (j!=i&&abs(a[i][j])>abs(a[p][q]))
{
p=i;q=j;
}
}
}
// 如果最大元充分小,結束
if (abs(a[p][q])<epsilon)
break;
lamda = 2*a[p][q];
mu = (a[p][p]-a[q][q]);
// printf("tan=%f\n",lamda/mu);
if (abs(mu)<epsilon)
{
w=1.;
}
else
{
if (mu>0)
w=lamda/sqrt(lamda*lamda+mu*mu);
else
w=-lamda/sqrt(lamda*lamda+mu*mu);
}
s=w/sqrt(2*(1+sqrt(1-w*w)));
c=sqrt(1-s*s);
// 做一次旋轉
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
b[i][j]=a[i][j];
}
// p行q行
for (j=0;j<n;j++)
{
if (j!=p&&j!=q)
{
b[p][j] = a[p][j]*c+a[q][j]*s;
}
if (j!=p&&j!=q)
{
b[q][j]= -a[p][j]*s+a[q][j]*c;
}
}
// p列q列
for (i=0;i<n;i++)
{
if (i!=p&&i!=q)
{
b[i][p] = a[i][p]*c+a[i][q]*s;
}
if (i!=p&&i!=q)
{
b[i][q]= -a[i][p]*s+a[i][q]*c;
}
}
// 對角元
b[p][p]=a[p][p]*c*c+a[q][q]*s*s+2*a[p][q]*s*c;
b[q][q]=a[p][p]*s*s+a[q][q]*c*c-2*a[p][q]*s*c;
b[p][q]=b[q][p] = 0.;
// b[p][q]=b[q][p] = (a[q][q]-a[p][p])*s*c+a[p][q]*(2*c*c-1);
// 特徵向量
for (i=0;i<n;i++) {
U[i][i]=1;
for (j=0;j<n;j++) {
if (j!=i)
{
U[i][j]=0;
}
}
}
U[p][p] = U[q][q]=c;
if (p<q)
{
U[p][q] = -s;
U[q][p] = s;
}
else
{
U[p][q] = s;
U[q][p] = -s;
}
for (i=0;i<n;i++) {
for (j=0;j<n;j++) {
Tmp[i][j] =0;
for (int k=0;k<n;k++) {
Tmp[i][j]+=B[i][k]*U[k][j];
}
}
}
k++;
// 如果旋轉後的值非常小, 置爲0
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
if (i!=j&&abs(b[i][j])<5e-10)
{
b[i][j] = 0;
}
a[i][j]=b[i][j];
B[i][j]=Tmp[i][j];
}
}
}
printf("a%d=\n",k);
for (i = 0; i < n; i++)
{
printf("%.14f\t",a[i][i]);
}
printf("\n");
printf("B%d=\n",k);
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
printf("%.14f\t",B[i][j]);
}
printf("\n");
}
}