作业打卡

研究Rosenbrock函数，其定义为 $f = (x_1-a)^{2} +(x_2-x_1^2)^2$

问题一：
a,b的选择会对函数的凹凸性会产生很大的变化，但是函数的大体形状不变。

问题二：求解最优解采用的最速下降法，初始点为(1,1).

import numpy as np
import matplotlib.pyplot as plt
def f(x1,x2,a,b):
    return (a-x1)**2 + b*(x2-x1**2)**2

def GetGradient(prex,a,b):
    x1 = prex[0]
    x2 = prex[1]
    return np.array([2*(x1-a)+4*b*x1*(x1**2-x2), 2*b*(x2-x1**2)])
def Getminpoint(x1,x2,a,b):
    mu = 0.01
    prex = np.array([x1,x2])
    prey = f(prex[0],prex[1],a,b)
    nowx = prex - mu*GetGradient(prex,a,b)
    nowy = f(nowx[0],nowx[1],a,b)
    epoch = 0
    while (nowy-prey)< 0.001 :
        prex =  nowx
        prey = nowy
        nowx = prex - mu*GetGradient(prex,a,b)
        nowy = f(nowx[0],nowx[1],a,b)
        epoch += 1
        if  epoch > 20:
            break
    return (nowx,nowy)

for a in [0,4,8,12]:
    for b in [-8,0,8,16]:
        fig = plt.figure()
        ax = plt.axes(projection='3d')
        minxy,minz = Getminpoint(1,1,a,b)
        ax.scatter(minxy[0],minxy[1],minz,c="red",zorder= 0,marker='o',s=100)

        x = np.linspace(minxy[0]-12,minxy[0]+12,30)
        y = np.linspace(minxy[1]-12,minxy[1]+12,30)
        X, Y = np.meshgrid(x, y)
        Z = f(X,Y,a,b)

       
        ax.plot_surface(X, Y, Z, cmap='viridis',zorder= 1)
        ax.set_xlabel('x')
        ax.set_ylabel('y')
        ax.set_zlabel('z')
        #调整观察角度和方位角。这里将俯仰角设为60度，把方位角调整为35度
        ax.view_init(60, 35)
        plt.savefig(r"C:\Users\huangjianguo\Desktop\SelfStudy\ensemble-learning-main\CH1-机器学习的数学基础\CH1-机器学习的数学基础\阶段一作业\a=%f,b=%f.jpg"%(a,b))
        print("%f,%f"%(a,b))