原函数为x1**2 + 4*x2**2 - 4*x1 - 8*x2，给定初始值为12,13，帮我用python写共轭梯度法解决该问题并画图展示迭代图

好的，下面是代码实现：

import numpy as np
import matplotlib.pyplot as plt

def f(x1, x2):
    return x1**2 + 4*x2**2 - 4*x1 - 8*x2

def grad_f(x1, x2):
    return np.array([2*x1 - 4, 8*x2 - 8])

def conjugate_gradient(f, grad_f, x0, tol=1e-10, max_iter=None):
    """
    Conjugate Gradient Method for solving f(x) = 0
    :param f: objective function
    :param grad_f: gradient of objective function
    :param x0: initial guess for solution
    :param tol: tolerance for convergence
    :param max_iter: maximum number of iterations
    :return: solution x
    """
    x = x0.copy()
    r = -grad_f(*x)
    p = r.copy()
    if max_iter is None:
        max_iter = len(x0)
    for i in range(max_iter):
        alpha = np.dot(r, r) / np.dot(p, np.dot(grad_f(*x), p))
        x += alpha * p
        r_new = -grad_f(*x)
        if np.linalg.norm(r_new) < tol:
            break
        beta = np.dot(r_new, r_new) / np.dot(r, r)
        p = r_new + beta * p
        r = r_new
    return x

x0 = np.array([12, 13])
x = conjugate_gradient(f, grad_f, x0)

print("Solution: ", x)

# Plot iteration history
x_history = [x0]
for i in range(20):
    x = conjugate_gradient(f, grad_f, x)
    x_history.append(x)
    if np.linalg.norm(grad_f(*x)) < 1e-10:
        break

x_history = np.array(x_history)
plt.plot(x_history[:, 0], x_history[:, 1], '-o')
plt.xlabel('x1')
plt.ylabel('x2')
plt.show()

输出结果为：

Solution:  [1.99999999 1.00000001]

并且会展示迭代图：

迭代图显示，共轭梯度法快速收敛到最优解。