Python实现最速下降法：基于SymPy与Backtracking Line Search

核心代码与Backtracking Line Search机制

from sympy import *
import numpy as np
def backtracking_line_search(f,df,x,x_k,p_k,alpha0):
  rho=0.5
  c=10**-4
  alpha=alpha0
  replacements1=zip(x,x_k)
  replacements2=zip(x,x_k+alpha*p_k)
  f_k=f.subs(replacements1)
  df_p=np.dot([df_.subs(replacements1) for df_ in df],p_k)
  while f.subs(replacements2)>f_k+c*alpha*df_p:
    alpha=rho*alpha
    replacements2 = zip(x, x_k +alpha * p_k)
  return alpha
def stepest_line_search(f,x,x0,alpha0):
  df = [diff(f, x_) for x_ in x]
  x_k=x0
  alpha=alpha0
  replacements=zip(x,x_k)
  len_df = sqrt(np.sum([df_.subs(replacements) ** 2 for df_ in df]))
  while len_df>1e-6:
    p_k=-1*np.array([df_.subs(replacements) for df_ in df])
    alpha = backtracking_line_search(f, df, x, x_k, p_k, alpha)
    x_k=x_k+alpha*p_k
    replacements = zip(x, x_k)
    len_df=np.sum([df_.subs(replacements)**2 for df_ in df])
  return x_k
if __name__=="__main__":
  init_printing(use_unicode=True)
  x1 = symbols("x1")
  x2 = symbols("x2")
  x = np.array([x1, x2])
  f = 100 * (x2 - x1 ** 2)**2 + (1 - x1) ** 2
  ans=stepest_line_search(f, x, np.array([1.2, 1]), 1)
  print "the minimal value in point:",ans