用Python实现最速下降法求极值的方法

'''
线性搜索子函数
'''

import numpy as np
import random

def goldsteinsearch(f,df,d,x,alpham,rho,t):

  flag=0

  a=0
  b=alpham
  fk=f(x)
  gk=df(x)

  phi0=fk
  dphi0=np.dot(gk,d)

  alpha=b*random.uniform(0,1)

  while(flag==0):
    newfk=f(x+alpha*d)
    phi=newfk
    if(phi-phi0<=rho*alpha*dphi0):
      if(phi-phi0>=(1-rho)*alpha*dphi0):
        flag=1
      else:
        a=alpha
        b=b
        if(b<alpham):
          alpha=(a+b)/2
        else:
          alpha=t*alpha
    else:
      a=a
      b=alpha
      alpha=(a+b)/2
  return alpha

"""
最速下降法
Rosenbrock函数
函数 f(x)=100*(x(2)-x(1).^2).^2+(1-x(1)).^2
梯度 g(x)=(-400*(x(2)-x(1)^2)*x(1)-2*(1-x(1)),200*(x(2)-x(1)^2))^(T)
"""

import numpy as np
import matplotlib.pyplot as plt
import random
import linesearch
from linesearch import goldsteinsearch

def rosenbrock(x):
  return 100*(x[1]-x[0]**2)**2+(1-x[0])**2

def jacobian(x):
  return np.array([-400*x[0]*(x[1]-x[0]**2)-2*(1-x[0]),200*(x[1]-x[0]**2)])


X1=np.arange(-1.5,1.5+0.05,0.05)
X2=np.arange(-3.5,2+0.05,0.05)
[x1,x2]=np.meshgrid(X1,X2)
f=100*(x2-x1**2)**2+(1-x1)**2; # 给定的函数
plt.contour(x1,x2,f,20) # 画出函数的20条轮廓线

def steepest(x0):

  print('初始点为:')
  print(x0,'\n')  
  imax = 20000
  W=np.zeros((2,imax))
  W[:,0] = x0
  i = 1   
  x = x0
  grad = jacobian(x)
  delta = sum(grad**2) # 初始误差


  while i<imax and delta>10**(-5):
    p = -jacobian(x)
    x0=x
    alpha = goldsteinsearch(rosenbrock,jacobian,p,x,1,0.1,2)
    x = x + alpha*p
    W[:,i] = x
    grad = jacobian(x)
    delta = sum(grad**2)
    i=i+1

  print("迭代次数为:",i)
  print("近似最优解为:")
  print(x,'\n')  
  W=W[:,0:i] # 记录迭代点
  return W

x0 = np.array([-1.2,1])
W=steepest(x0)

plt.plot(W[0,:],W[1,:],'g*',W[0,:],W[1,:]) # 画出迭代点收敛的轨迹
plt.show()

import linesearch
from linesearch import goldsteinsearch

初始点为:

[-1.2 1. ] 

迭代次数为: 1504

近似最优解为:

[ 1.00318532 1.00639618]

迭代点的轨迹为 

初始点为:
[-1.2 1. ] 

迭代次数为: 1994

近似最优解为:
[ 0.99735222 0.99469882]