2018-10-02 21:24:06 +08:00
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''' mbinary
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#########################################################################
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# File : iteration.py
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# Author: mbinary
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# Mail: zhuheqin1@gmail.com
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2019-01-31 12:09:46 +08:00
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# Blog: https://mbinary.xyz
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2018-10-02 21:24:06 +08:00
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# Github: https://github.com/mbinary
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# Created Time: 2018-10-02 21:14
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# Description:
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#########################################################################
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'''
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import sympy
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import numpy as np
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from math import sqrt
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def newton(y:sympy.core,x0:float,epsilon:float=0.00001,maxtime:int=50) ->(list,list):
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'''
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newton 's iteration method for finding a zeropoint of a func
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y is the func, x0 is the init x val: int float epsilon is the accurrency
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'''
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if epsilon <0:epsilon = -epsilon
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ct =0
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t = y.free_symbols
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varsymbol = 'x' if len(t)==0 else t.pop()
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x0= float(x0)
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y_diff = y.diff()
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li = [x0]
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vals = []
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while 1:
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val = y.subs(varsymbol,x0)
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vals.append(val)
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x = x0- val/y_diff.subs(varsymbol,x0)
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li.append(x)
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ct+=1
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if ct>maxtime:
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print("after iteration for {} times, I still havn't reach the accurrency.\
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Maybe this function havsn't zeropoint\n".format(ct))
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return li ,val
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if abs(x-x0)<epsilon:return li,vals
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x0 = x
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def secant(y:sympy.core,x0:float,x1:float,epsilon:float =0.00001,maxtime:int=50) ->(list,list):
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'''
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弦截法, 使用newton 差商计算,每次只需计算一次f(x)
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secant method for finding a zeropoint of a func
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y is the func , x0 is the init x val, epsilon is the accurrency
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'''
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if epsilon <0:epsilon = -epsilon
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ct =0
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x0,x1 = float(x0),float(x1)
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li = [x0,x1]
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t = y.free_symbols
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varsymbol = 'x' if len(t)==0 else t.pop()
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last = y.subs(varsymbol,x0)
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vals = [last]
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while 1:
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cur = y.subs(varsymbol,x1)
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vals.append(cur)
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x = x1-cur*(x1-x0)/(cur-last)
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x0 ,x1= x1,x
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last = cur
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li.append(x)
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ct+=1
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if ct>maxtime:
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print("after iteration for {} times, I still havn't reach the accurrency.\
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Maybe this function havsn't zeropoint\n".format(ct))
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return li,vals
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if abs(x0-x1)<epsilon:return li,vals
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x0 = x
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def solveNonlinearEquations(funcs:[sympy.core],init_dic:dict,epsilon:float=0.001,maxtime:int=50)->dict:
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'''solve nonlinear equations:'''
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li = list(init_dic.keys())
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delta = {i:0 for i in li}
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ct = 0
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while 1:
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ys = np.array([f.subs(init_dic) for f in funcs],dtype = 'float')
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mat = np.matrix([[i.diff(x).subs(init_dic) for x in li] for i in funcs ],dtype = 'float')
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delt = np.linalg.solve(mat,-ys)
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for i,j in enumerate(delt):
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init_dic[li[i]] +=j
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delta[li[i]] = j
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if ct>maxtime:
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print("after iteration for {} times, I still havn't reach the accurrency.\
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Maybe this function havsn't zeropoint\n".format(ct))
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return init_dic
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if sqrt(sum(i**2 for i in delta.values()))<epsilon:return init_dic
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if __name__ =='__main__':
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x,y,z = sympy.symbols('x y z')
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res,res2= newton(x**5-9,2,0.01)
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print(res,res2)
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res,res2 = secant (x**3-3*x-2,1,3,1e-3)
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print(res,res2)
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funcs=[x**2+y**2-1,x**3-y]
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init = {x:0.8,y:0.6}
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res_dic = solveNonlinearEquations(funcs,init,0.001)
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print(res_dic)
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