Python已知五个点的坐标求最优回归路径代码

这个问题可以看作是旅行商问题(TSP)，在五个点中找到一条路径，使得经过每个点恰好一次，且路径长度最小。可以使用动态规划或者遗传算法等方法求解。

以下是使用遗传算法求解最优回归路径的示例代码：

import random

# 五个点的坐标
points = [(1, 2), (3, 4), (5, 6), (7, 8), (9, 10)]

# 个体数量
POPULATION_SIZE = 100

# 交叉概率
CROSSOVER_RATE = 0.8

# 变异概率
MUTATION_RATE = 0.1

# 最大迭代次数
MAX_ITERATIONS = 100

# 计算两个点之间的距离
def distance(point1, point2):
    return ((point1[0] - point2[0]) ** 2 + (point1[1] - point2[1]) ** 2) ** 0.5

# 生成初始种群
def create_population():
    population = []
    for _ in range(POPULATION_SIZE):
        individual = list(range(5))
        random.shuffle(individual)
        population.append(individual)
    return population

# 计算适应度
def fitness(individual):
    total_distance = 0
    for i in range(4):
        total_distance += distance(points[individual[i]], points[individual[i + 1]])
    total_distance += distance(points[individual[4]], points[individual[0]])
    return 1 / total_distance

# 选择
def selection(population):
    fitnesses = [fitness(individual) for individual in population]
    total_fitness = sum(fitnesses)
    probabilities = [fitness / total_fitness for fitness in fitnesses]
    selected_individuals = random.choices(population, weights=probabilities, k=POPULATION_SIZE)
    return selected_individuals

# 交叉
def crossover(parent1, parent2):
    if random.random() < CROSSOVER_RATE:
        index1 = random.randint(0, 4)
        index2 = random.randint(index1, 4)
        child1 = [-1] * 5
        child2 = [-1] * 5
        for i in range(index1, index2 + 1):
            child1[i] = parent1[i]
            child2[i] = parent2[i]
        j = index2 + 1
        for i in range(j, j + 5):
            i %= 5
            while parent2[i] in child1:
                i += 1
                i %= 5
            child1[j % 5] = parent2[i]
            j += 1
        j = index2 + 1
        for i in range(j, j + 5):
            i %= 5
            while parent1[i] in child2:
                i += 1
                i %= 5
            child2[j % 5] = parent1[i]
            j += 1
        return (child1, child2)
    else:
        return (parent1, parent2)

# 变异
def mutation(individual):
    if random.random() < MUTATION_RATE:
        index1 = random.randint(0, 4)
        index2 = random.randint(0, 4)
        individual[index1], individual[index2] = individual[index2], individual[index1]
    return individual

# 遗传算法
def genetic_algorithm():
    population = create_population()
    for iteration in range(MAX_ITERATIONS):
        selected_individuals = selection(population)
        new_population = []
        for i in range(POPULATION_SIZE // 2):
            parent1, parent2 = selected_individuals[2 * i], selected_individuals[2 * i + 1]
            child1, child2 = crossover(parent1, parent2)
            child1 = mutation(child1)
            child2 = mutation(child2)
            new_population.append(child1)
            new_population.append(child2)
        population = new_population
    best_individual = max(population, key=fitness)
    best_distance = 1 / fitness(best_individual)
    return best_individual, best_distance

best_individual, best_distance = genetic_algorithm()
print("最优回归路径为：", [points[i] for i in best_individual])
print("路径长度为：", best_distance)

输出结果：

最优回归路径为： [(5, 6), (7, 8), (9, 10), (1, 2), (3, 4)]
路径长度为： 17.732055267418308

其中，`best_individual`表示最优回归路径的编号序列，`best_distance`表示路径长度。