编写程序用人工免疫算法求解函数f(x)=x1^2+x2^2-5x1x2-2x1-4x2+10的极值

非常感谢您的提问，我可以回答这个编程类的问题。以下是用人工免疫算法求解函数f(x)=x1^2 x2^2-5x1x2-2x1-4x2 10的极值的程序代码：

import numpy as np

def f(x):
    return x[0]**2 * x[1]**2 - 5*x[0]*x[1] - 2*x[0] - 4*x[1] + 10

def immune_algorithm(f, bounds, pop_size=50, max_iter=1000, beta=0.1, gamma=0.1, mu=0.1):
    dim = len(bounds)
    pop = np.random.uniform(bounds[:, 0], bounds[:, 1], size=(pop_size, dim))
    fitness = np.array([f(p) for p in pop])
    best_idx = np.argmin(fitness)
    best = pop[best_idx]
    for i in range(max_iter):
        # clone
        num_clones = np.round(beta * pop_size).astype(int)
        clones = np.repeat(pop, num_clones, axis=0)
        # hypermutation
        clones += np.random.normal(0, gamma, size=clones.shape)
        clones = np.clip(clones, bounds[:, 0], bounds[:, 1])
        # selection
        clone_fitness = np.array([f(p) for p in clones])
        sort_idx = np.argsort(clone_fitness)
        pop = clones[sort_idx[:pop_size]]
        fitness = clone_fitness[sort_idx[:pop_size]]
        # mutation
        num_mutants = np.round(mu * pop_size).astype(int)
        mutants = np.random.uniform(bounds[:, 0], bounds[:, 1], size=(num_mutants, dim))
        mutant_fitness = np.array([f(p) for p in mutants])
        # replace worst individuals
        sort_idx = np.argsort(fitness)
        pop[sort_idx[-num_mutants:]] = mutants[mutant_fitness > fitness[sort_idx[-num_mutants:]]]
        fitness[sort_idx[-num_mutants:]] = mutant_fitness[mutant_fitness > fitness[sort_idx[-num_mutants:]]]
        # update best
        best_idx = np.argmin(fitness)
        if fitness[best_idx] < f(best):
            best = pop[best_idx]
    return best, f(best)

bounds = np.array([[-5, 5], [-5, 5]])
best, value = immune_algorithm(f, bounds)
print("Best solution:", best)
print("Best value:", value)

这个程序使用了人工免疫算法来求解函数f(x)的极值。算法的基本思想是通过克隆、超突变和选择等操作来搜索解空间，并通过突变来保持种群的多样性。最终，算法返回找到的最优解及其对应的函数值。

请注意，这个程序仅供参考，实际应用中可能需要根据具体问题进行调整和优化。