# -*- coding: utf-8 -*-
"""
Created on Tue Jun  8 17:19:05 2021

@author: hzh
"""
# 例2：求函数y=x2-10x-30  的最小值
# 绘图
import numpy as np
import matplotlib.pyplot as plt


def f(x):
	return x ** 2 - 10 * x - 30


x = np.arange(-10, 10, 0.1)
plt.plot(x, f(x))
plt.grid()
plt.show()

# 梯度下降法求函数最值
np.random.seed(3)
x = np.random.randn(1);  # 10代表数据的维度
# print(x)
learning_rate = 0.1
err = 0.000001;
max_iters = 10000


def f(x):
	return x ** 2 - 10 * x - 30


def df(x):
	return 2 * x - 10


for i in range(max_iters):
	print("第 %d 次迭代：x=%.8f y=%.8f" % (i, x, f(x)))
	if abs(df(x)) < err:
		break
	x = x - df(x) * learning_rate  # xk+1=xk- η* f’(xk)  (迭代公式)

# 练习题1：求函数y=x2-8x-25的最小值
# 练习题2：
# 已知函数y=x+sqrt(2*R*x-x**2), R为大于0的常数，x的取值范围在[R,2*R]之间。
# 当R=10时，求该函数在x取值区间内函数的最大值。
# 绘图

# 梯度下降法求函数最大值


# 给定一个二元函数f(x,y)= - exp(x-y)*(x**2-2*y**2)，
# 用梯度下降法求其在x<0,y<0范围上的极小值

x = -1;
y = -1
learning_rate = 0.1
err = 0.000001;
max_iters = 10000


def f(x, y):
	return -np.exp(x - y) * (x ** 2 - 2 * y ** 2)


def dx(x, y):
	return -(np.exp(x - y) * (2 * x) + (x ** 2 - 2 * y ** 2) * np.exp(x - y))


def dy(x, y):
	return -(np.exp(x - y) * (-4 * y) + (x ** 2 - 2 * y ** 2) * np.exp(x - y) * (-1))


for t in range(max_iters):
	if t % 100 == 0:
		print("Iter %d, x=%.8f,y=%.8f,z=%.8f,dx=%.8f,dy=%.8f" % (t, x, y, f(x, y), dx(x, y), dy(x, y)))
	if abs(dx(x, y)) < err and abs(dy(x, y)) < err:
		print("Iter %d, x=%.8f,y=%.8f,z=%.8f,dx=%.8f,dy=%.8f" % (t, x, y, f(x, y), dx(x, y), dy(x, y)))
		break
	x = x - learning_rate * dx(x, y);
	y = y - learning_rate * dy(x, y)  # 迭代公式

'''#练习题：
二元函数f(x,y)=x3-y3+3x2+3y2-9x求最值
三元函数f(x,y,z)=x2+2y2+3z2+2x+4y-6z求最值
'''