假设有一块材料模板,长宽确定(分别为3000,1500),有两种切割形状(可是更多切割形状),切割形状p1为长度(373,201),切割形状p2为长度(406,229),那么请问两者切割数量以及最后的切割位置(保证原材料木板利用率最高的情况下)?用遗传算法解答(包含代码)
时间: 2024-01-30 15:02:27 浏览: 203
好的,我可以帮你提供一个基本的遗传算法的实现代码来解决这个问题。请注意,这只是一个简化的示例代码,实际应用中可能需要根据具体情况进行调整和优化。
```python
import random
# 定义问题参数
material_width = 3000
material_height = 1500
shapes = {
'p1': {'width': 373, 'height': 201},
'p2': {'width': 406, 'height': 229}
}
# 定义遗传算法参数
population_size = 50
generations = 100
mutation_rate = 0.01
# 初始化种群
def initialize_population():
population = []
for _ in range(population_size):
individual = []
for _ in range(len(shapes)):
individual.append(random.randint(0, 1))
population.append(individual)
return population
# 计算适应度函数(原材料利用率)
def calculate_fitness(individual):
total_width = 0
total_height = 0
for i, shape in enumerate(shapes.values()):
if individual[i] == 1:
total_width += shape['width']
total_height += shape['height']
utilization = (total_width * total_height) / (material_width * material_height)
return utilization
# 选择操作
def selection(population):
sorted_population = sorted(population, key=lambda x: calculate_fitness(x), reverse=True)
return sorted_population[:int(len(population)/2)]
# 交叉操作
def crossover(parent1, parent2):
crossover_point = random.randint(1, len(shapes)-1)
child1 = parent1[:crossover_point] + parent2[crossover_point:]
child2 = parent2[:crossover_point] + parent1[crossover_point:]
return child1, child2
# 变异操作
def mutation(individual):
for i in range(len(individual)):
if random.random() < mutation_rate:
individual[i] = 1 - individual[i]
return individual
# 遗传算法主程序
def genetic_algorithm():
population = initialize_population()
for generation in range(generations):
new_population = []
# 选择操作
selected_population = selection(population)
# 交叉操作
for _ in range(population_size):
parent1 = random.choice(selected_population)
parent2 = random.choice(selected_population)
child1, child2 = crossover(parent1, parent2)
new_population.append(child1)
new_population.append(child2)
# 变异操作
for individual in new_population:
mutated_individual = mutation(individual)
population.append(mutated_individual)
# 更新种群
population = population[:population_size]
best_individual = max(population, key=lambda x: calculate_fitness(x))
best_fitness = calculate_fitness(best_individual)
return best_individual, best_fitness
# 运行遗传算法
best_individual, best_fitness = genetic_algorithm()
# 输出结果
print("最佳解:", best_individual)
print("原材料利用率:", best_fitness)
```
请注意,以上代码仅提供了一个基本的遗传算法框架,具体的实现可能需要根据实际情况进行调整和优化。同时,由于问题描述较为简化,可能无法保证找到最优解。在实际应用中,可能需要采用更复杂的算法或者进行进一步的问题建模和约束处理。
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