1/3x^2+1/2y^2实现带armijo梯度下降法代码并显示轨迹图

以下是使用Armijo梯度下降法来最小化函数1/3x^2+1/2y^2的代码，并绘制出轨迹图。这里我们使用Python和matplotlib库来编写代码。

import numpy as np
import matplotlib.pyplot as plt

# 定义函数f(x,y) = 1/3x^2 + 1/2y^2
def f(x, y):
    return 1/3 * x**2 + 1/2 * y**2

# 定义梯度函数grad(x,y)
def grad(x, y):
    return np.array([2/3 * x, y])

# 定义Armijo梯度下降法函数armijo(x, y, alpha, beta)
def armijo(x, y, alpha, beta):
    # 初始化参数
    x_list = [x]
    y_list = [y]
    alpha_list = [alpha]
    beta_list = [beta]
    epsilon = 1e-8
    max_iter = 10000
    iter_num = 0
    while True:
        # 计算梯度
        g = grad(x, y)
        # 计算步长
        t = 1
        while f(x - t * g[0], y - t * g[1]) > f(x, y) - alpha * t * np.dot(g, g):
            t *= beta
        # 更新参数
        x = x - t * g[0]
        y = y - t * g[1]
        # 记录参数和参数变化情况
        x_list.append(x)
        y_list.append(y)
        alpha_list.append(t)
        beta_list.append(beta)
        # 判断是否收敛
        if np.linalg.norm(grad(x, y)) < epsilon or iter_num > max_iter:
            break
        iter_num += 1
    return x_list, y_list, alpha_list, beta_list

# 运行Armijo梯度下降法函数
x_list, y_list, alpha_list, beta_list = armijo(2, 2, 0.25, 0.5)

# 绘制轨迹图
fig, ax = plt.subplots()
ax.set_aspect('equal')
x = np.linspace(-3, 3, 100)
y = np.linspace(-3, 3, 100)
X, Y = np.meshgrid(x, y)
Z = f(X, Y)
ax.contour(X, Y, Z, 50)
ax.plot(x_list, y_list, 'r.-')
ax.set_xlabel('x')
ax.set_ylabel('y')
plt.show()

运行代码后，将会显示出函数的轮廓线以及最小化函数的轨迹。可以根据需要调整初始参数和Armijo梯度下降法的参数来探索更多不同的轨迹。