algorithm-in-python/math/numericalAnalysis/iteration.py

109 lines
3.3 KiB
Python

''' mbinary
#########################################################################
# File : iteration.py
# Author: mbinary
# Mail: zhuheqin1@gmail.com
# Blog: https://mbinary.coding.me
# Github: https://github.com/mbinary
# Created Time: 2018-10-02 21:14
# Description:
#########################################################################
'''
import sympy
import numpy as np
from math import sqrt
def newton(y:sympy.core,x0:float,epsilon:float=0.00001,maxtime:int=50) ->(list,list):
'''
newton 's iteration method for finding a zeropoint of a func
y is the func, x0 is the init x val: int float epsilon is the accurrency
'''
if epsilon <0:epsilon = -epsilon
ct =0
t = y.free_symbols
varsymbol = 'x' if len(t)==0 else t.pop()
x0= float(x0)
y_diff = y.diff()
li = [x0]
vals = []
while 1:
val = y.subs(varsymbol,x0)
vals.append(val)
x = x0- val/y_diff.subs(varsymbol,x0)
li.append(x)
ct+=1
if ct>maxtime:
print("after iteration for {} times, I still havn't reach the accurrency.\
Maybe this function havsn't zeropoint\n".format(ct))
return li ,val
if abs(x-x0)<epsilon:return li,vals
x0 = x
def secant(y:sympy.core,x0:float,x1:float,epsilon:float =0.00001,maxtime:int=50) ->(list,list):
'''
弦截法, 使用newton 差商计算,每次只需计算一次f(x)
secant method for finding a zeropoint of a func
y is the func , x0 is the init x val, epsilon is the accurrency
'''
if epsilon <0:epsilon = -epsilon
ct =0
x0,x1 = float(x0),float(x1)
li = [x0,x1]
t = y.free_symbols
varsymbol = 'x' if len(t)==0 else t.pop()
last = y.subs(varsymbol,x0)
vals = [last]
while 1:
cur = y.subs(varsymbol,x1)
vals.append(cur)
x = x1-cur*(x1-x0)/(cur-last)
x0 ,x1= x1,x
last = cur
li.append(x)
ct+=1
if ct>maxtime:
print("after iteration for {} times, I still havn't reach the accurrency.\
Maybe this function havsn't zeropoint\n".format(ct))
return li,vals
if abs(x0-x1)<epsilon:return li,vals
x0 = x
def solveNonlinearEquations(funcs:[sympy.core],init_dic:dict,epsilon:float=0.001,maxtime:int=50)->dict:
'''solve nonlinear equations:'''
li = list(init_dic.keys())
delta = {i:0 for i in li}
ct = 0
while 1:
ys = np.array([f.subs(init_dic) for f in funcs],dtype = 'float')
mat = np.matrix([[i.diff(x).subs(init_dic) for x in li] for i in funcs ],dtype = 'float')
delt = np.linalg.solve(mat,-ys)
for i,j in enumerate(delt):
init_dic[li[i]] +=j
delta[li[i]] = j
if ct>maxtime:
print("after iteration for {} times, I still havn't reach the accurrency.\
Maybe this function havsn't zeropoint\n".format(ct))
return init_dic
if sqrt(sum(i**2 for i in delta.values()))<epsilon:return init_dic
if __name__ =='__main__':
x,y,z = sympy.symbols('x y z')
res,res2= newton(x**5-9,2,0.01)
print(res,res2)
res,res2 = secant (x**3-3*x-2,1,3,1e-3)
print(res,res2)
funcs=[x**2+y**2-1,x**3-y]
init = {x:0.8,y:0.6}
res_dic = solveNonlinearEquations(funcs,init,0.001)
print(res_dic)