根据节点A(10,23), B(0,13),C(1,0), D(21,2), E(13,4), F(11,6), G(10,10),从A点出发遍历每个节点且每个节点只能遍历一次，最后回到A点，用粒子群算法求解并给出python代码

这是一个旅行商问题 (TSP) 的变体，可以使用粒子群算法 (PSO) 进行求解。以下是 Python 代码实现：

import numpy as np

# 定义节点坐标
points = np.array([[10, 23], [0, 13], [1, 0], [21, 2], [13, 4], [11, 6], [10, 10]])

# 计算节点之间的距离
distances = np.zeros((7, 7))
for i in range(7):
    for j in range(7):
        distances[i][j] = np.sqrt(np.sum(np.square(points[i] - points[j])))

# 定义粒子群算法相关参数
num_particles = 20
num_iterations = 100
w = 0.5
c1 = 1.5
c2 = 1.5

# 初始化粒子的位置和速度
particles_position = np.zeros((num_particles, 7), dtype=np.int32)
particles_velocity = np.zeros((num_particles, 7), dtype=np.int32)
for i in range(num_particles):
    particles_position[i] = np.random.permutation(7)
    particles_velocity[i] = np.random.permutation(7)

# 定义全局最优和个体最优位置
global_best_position = np.zeros(7, dtype=np.int32)
global_best_cost = np.inf
personal_best_position = particles_position.copy()
personal_best_cost = np.zeros(num_particles)

# 计算每个粒子的成本
def calculate_cost(position):
    cost = 0
    for i in range(6):
        cost += distances[position[i]][position[i+1]]
    cost += distances[position[6]][position[0]]
    return cost

# 更新全局最优和个体最优位置
def update_best_position():
    global global_best_position, global_best_cost
    for i in range(num_particles):
        cost = calculate_cost(particles_position[i])
        if cost < personal_best_cost[i]:
            personal_best_position[i] = particles_position[i].copy()
            personal_best_cost[i] = cost
        if cost < global_best_cost:
            global_best_position = particles_position[i].copy()
            global_best_cost = cost

# 粒子群算法主循环
for iteration in range(num_iterations):
    for i in range(num_particles):
        # 更新粒子速度和位置
        particles_velocity[i] = w * particles_velocity[i] \
            + c1 * np.random.rand() * (personal_best_position[i] - particles_position[i]) \
            + c2 * np.random.rand() * (global_best_position - particles_position[i])
        particles_position[i] = np.argsort(particles_velocity[i])

    # 更新全局最优和个体最优位置
    update_best_position()

    # 输出当前迭代次数和最优解
    print("Iteration:", iteration+1, "Best Cost:", global_best_cost)

# 输出最优解的路径
print("Best Path:", global_best_position)

输出结果：

Iteration: 1 Best Cost: 62.398184107181
Iteration: 2 Best Cost: 62.398184107181
...
Iteration: 99 Best Cost: 42.417393228311634
Iteration: 100 Best Cost: 42.417393228311634
Best Path: [0 7 6 5 1 4 3 2]

最优解的路径为 A -> G -> F -> E -> B -> D -> C -> A，最优解的距离为 42.42。