组合太多

【问题标题】：Too many combinations组合太多
【发布时间】：2021-02-01 11:59:14
【问题描述】：

您好，我正在尝试生成所有可能的工人与建筑物的组合。 （让我解释一下我的情况）：

我在我的世界上玩MineColonies。在这个模组中，你有可以在建筑物中分配工作的殖民者。这些工人有技能和分配给他们的分数。 （如敏捷：20，力量：5 等），并且当分配一个技能与之相称的殖民者时，建筑物的工作会表现得更好......

所以我创建了一个包含所有工人和建筑物的数据库，并希望优化哪些工人在哪些建筑物工作。

buildings_dict = {1: ['Strength', 'Focus'],
                  2: ['Adaptability', 'Athletics'],
                  3: ['Knowledge', 'Dexterity'],
                  4: ['Adaptability', 'Knowledge'],
                  6: ['Stamina', 'Athletics'],
                  5: ['Athletics', 'Stamina'],
                  7: ['Focus', 'Agility'],
                  8: ['Dexterity', 'Creativity'],
                  9: ['Strength', 'Focus'],
                  10: ['Adaptability', 'Stamina'],
                  11: ['Agility', 'Adaptability'],
                  12: ['Mana', 'Knowledge'],
                  13: ['Strength', 'Stamina'],
                  14: ['Athletics', 'Strength'],
                  15: ['Creativity', 'Dexterity'],
                  16: ['Knowledge', 'Mana'],
                  17: ['Agility', 'Adaptability']}

workers_dict = {3: {'Mana': 30,
  'Focus': 1,
  'Agility': 3,
  'Stamina': 3,
  'Knowlege': 30,
  'Strenght': 8,
  'Athletics': 13,
  'Dexterity': 6,
  'Creativity': 10,
  'Adaptability': 10,
  'Intelligence': 10},
 4: {'Mana': 29,
  'Focus': 32,
  'Agility': 22,
  'Stamina': 28,
  'Knowlege': 21,
  'Strenght': 30,
  'Athletics': 20,
  'Dexterity': 31,
  'Creativity': 31,
  'Adaptability': 8,
  'Intelligence': 18},
 5: {'Mana': 13,
  'Focus': 1,
  'Agility': 9,
  'Stamina': 27,
  'Knowlege': 9,
  'Strenght': 13,
  'Athletics': 15,
  'Dexterity': 21,
  'Creativity': 16,
  'Adaptability': 13,
  'Intelligence': 28},
 6: {'Mana': 17,
  'Focus': 14,
  'Agility': 10,
  'Stamina': 17,
  'Knowlege': 13,
  'Strenght': 5,
  'Athletics': 10,
  'Dexterity': 15,
  'Creativity': 1,
  'Adaptability': 11,
  'Intelligence': 4},
 7: {'Mana': 1,
  'Focus': 8,
  'Agility': 6,
  'Stamina': 27,
  'Knowlege': 11,
  'Strenght': 17,
  'Athletics': 30,
  'Dexterity': 1,
  'Creativity': 5,
  'Adaptability': 11,
  'Intelligence': 5},
 8: {'Mana': 6,
  'Focus': 1,
  'Agility': 12,
  'Stamina': 30,
  'Knowlege': 20,
  'Strenght': 15,
  'Athletics': 30,
  'Dexterity': 9,
  'Creativity': 17,
  'Adaptability': 30,
  'Intelligence': 19},
 9: {'Mana': 5,
  'Focus': 7,
  'Agility': 19,
  'Stamina': 5,
  'Knowlege': 22,
  'Strenght': 18,
  'Athletics': 26,
  'Dexterity': 10,
  'Creativity': 24,
  'Adaptability': 20,
  'Intelligence': 22},
 10: {'Mana': 8,
  'Focus': 12,
  'Agility': 27,
  'Stamina': 3,
  'Knowlege': 17,
  'Strenght': 1,
  'Athletics': 5,
  'Dexterity': 9,
  'Creativity': 7,
  'Adaptability': 29,
  'Intelligence': 1},
 11: {'Mana': 1,
  'Focus': 4,
  'Agility': 5,
  'Stamina': 30,
  'Knowlege': 16,
  'Strenght': 11,
  'Athletics': 28,
  'Dexterity': 11,
  'Creativity': 5,
  'Adaptability': 12,
  'Intelligence': 4},
 12: {'Mana': 7,
  'Focus': 1,
  'Agility': 17,
  'Stamina': 25,
  'Knowlege': 23,
  'Strenght': 4,
  'Athletics': 8,
  'Dexterity': 26,
  'Creativity': 15,
  'Adaptability': 29,
  'Intelligence': 22},
 13: {'Mana': 2,
  'Focus': 1,
  'Agility': 5,
  'Stamina': 21,
  'Knowlege': 24,
  'Strenght': 18,
  'Athletics': 20,
  'Dexterity': 10,
  'Creativity': 12,
  'Adaptability': 30,
  'Intelligence': 5},
 14: {'Mana': 9,
  'Focus': 16,
  'Agility': 14,
  'Stamina': 25,
  'Knowlege': 14,
  'Strenght': 24,
  'Athletics': 30,
  'Dexterity': 9,
  'Creativity': 19,
  'Adaptability': 23,
  'Intelligence': 18},
 15: {'Mana': 23,
  'Focus': 15,
  'Agility': 5,
  'Stamina': 12,
  'Knowlege': 24,
  'Strenght': 12,
  'Athletics': 20,
  'Dexterity': 29,
  'Creativity': 5,
  'Adaptability': 19,
  'Intelligence': 12},
 17: {'Mana': 21,
  'Focus': 23,
  'Agility': 30,
  'Stamina': 18,
  'Knowlege': 27,
  'Strenght': 7,
  'Athletics': 30,
  'Dexterity': 10,
  'Creativity': 5,
  'Adaptability': 22,
  'Intelligence': 18},
 18: {'Mana': 11,
  'Focus': 11,
  'Agility': 4,
  'Stamina': 7,
  'Knowlege': 28,
  'Strenght': 11,
  'Athletics': 20,
  'Dexterity': 28,
  'Creativity': 13,
  'Adaptability': 12,
  'Intelligence': 30},
 19: {'Mana': 11,
  'Focus': 11,
  'Agility': 4,
  'Stamina': 7,
  'Knowlege': 28,
  'Strenght': 11,
  'Athletics': 20,
  'Dexterity': 28,
  'Creativity': 13,
  'Adaptability': 12,
  'Intelligence': 30},
 20: {'Mana': 15,
  'Focus': 20,
  'Agility': 28,
  'Stamina': 22,
  'Knowlege': 18,
  'Strenght': 15,
  'Athletics': 23,
  'Dexterity': 19,
  'Creativity': 20,
  'Adaptability': 27,
  'Intelligence': 20},
 21: {'Mana': 30,
  'Focus': 7,
  'Agility': 9,
  'Stamina': 7,
  'Knowlege': 30,
  'Strenght': 3,
  'Athletics': 6,
  'Dexterity': 17,
  'Creativity': 4,
  'Adaptability': 11,
  'Intelligence': 28},
 22: {'Mana': 9,
  'Focus': 10,
  'Agility': 28,
  'Stamina': 26,
  'Knowlege': 1,
  'Strenght': 8,
  'Athletics': 5,
  'Dexterity': 26,
  'Creativity': 1,
  'Adaptability': 14,
  'Intelligence': 16},
 23: {'Mana': 4,
  'Focus': 14,
  'Agility': 19,
  'Stamina': 5,
  'Knowledge': 21,
  'Strength': 25,
  'Athletics': 12,
  'Dexterity': 23,
  'Creativity': 26,
  'Adaptability': 21,
  'Intelligence': 22},
 24: {'Mana': 1,
  'Focus': 1,
  'Agility': 18,
  'Stamina': 24,
  'Knowledge': 25,
  'Strength': 20,
  'Athletics': 9,
  'Dexterity': 14,
  'Creativity': 19,
  'Adaptability': 30,
  'Intelligence': 7},
 25: {'Mana': 12,
  'Focus': 13,
  'Agility': 21,
  'Stamina': 23,
  'Knowledge': 11,
  'Strength': 16,
  'Athletics': 18,
  'Dexterity': 24,
  'Creativity': 1,
  'Adaptability': 20,
  'Intelligence': 1},
 26: {'Mana': 10,
  'Focus': 14,
  'Agility': 12,
  'Stamina': 27,
  'Knowledge': 17,
  'Strength': 24,
  'Athletics': 23,
  'Dexterity': 21,
  'Creativity': 5,
  'Adaptability': 5,
  'Intelligence': 28},
 27: {'Mana': 11,
  'Focus': 23,
  'Agility': 21,
  'Stamina': 12,
  'Knowledge': 15,
  'Strength': 24,
  'Athletics': 17,
  'Dexterity': 12,
  'Creativity': 1,
  'Adaptability': 11,
  'Intelligence': 9},
 28: {'Mana': 7,
  'Focus': 21,
  'Agility': 22,
  'Stamina': 21,
  'Knowledge': 14,
  'Strength': 15,
  'Athletics': 9,
  'Dexterity': 16,
  'Creativity': 2,
  'Adaptability': 11,
  'Intelligence': 5},
 29: {'Mana': 12,
  'Focus': 25,
  'Agility': 29,
  'Stamina': 6,
  'Knowledge': 7,
  'Strength': 10,
  'Athletics': 14,
  'Dexterity': 15,
  'Creativity': 6,
  'Adaptability': 13,
  'Intelligence': 29},
 30: {'Mana': 21,
  'Focus': 17,
  'Agility': 8,
  'Stamina': 21,
  'Knowledge': 22,
  'Strength': 22,
  'Athletics': 26,
  'Dexterity': 13,
  'Creativity': 15,
  'Adaptability': 24,
  'Intelligence': 13}}

抱歉，代码块太长了，是的，我意识到 id 不一定正确（想让它可重现）。

所以我使用itertools.permutations 将所有工人组合带到建筑物：

import itertools
workers_ls = list(workers_dict.keys())
combinations = list(itertools.permutations(workers_ls, len(buildings_dict))

（我打算之后给组合打分）

这显然从未完成运行，因为它大约是 27 ！ = 1×10²⁸。 我想知道我的问题是否有另一种解决方案，或者是否有一种方法可以在不经过所有组合的情况下确定最佳解决方案。 （我愿意用其他编码语言工作）

谢谢！

【问题讨论】：

    标签： python combinations permutation itertools largenumber


    【解决方案1】：

    我假设您希望最大化总产量的总和。例如，当没有分配工人时，总产量为零（或不依赖于工人分配的某个常数）。如果您将工人与Agility 2 和Focus 3 与具有[Agility, Focus] 属性的建筑物配对，则将2+3=5 添加到总产量中。

    此类问题通常通过线性规划来解决。我将使用pulp 来帮助制定线性规划问题。我还建议查看Julia 包JuMP。

    计算总产量的实际规则可能更复杂。如果 (1) 您可以定义一些生产矩阵的模拟，并且 (2) 总生产可以表示为（工人、建筑）对的生产之和，则您仍然可以使用线性规划。

    这里有两种方法可以解决这个问题。第一个允许每个建筑物有多个工人，第二个不允许。

    设置

    import pandas as pd
import numpy as np
# !pip install pulp
import pulp

df_buildings = pd.DataFrame(buildings_dict).T
df_workers = pd.DataFrame(workers_dict).T

# there are a few typos, e.g. Strenght vs. Strength and Knowlege vs. Knowledge
# let's fix this first
df_workers.Knowledge.fillna(df_workers.Knowlege, inplace=True)
df_workers.Strength.fillna(df_workers.Strenght, inplace=True)
del df_workers["Strenght"], df_workers["Knowlege"]

# fixing some notation
workers = df_workers.index.tolist() # list of workers
buildings =  df_buildings.index.tolist() # list of building

# next, we define production matrix
# production[i,j] will contain the productivity of 
# worker i when assigned to building j
# you could vectorize this step, though it seems fast enough here
production = pd.DataFrame(index=workers, columns=buildings)
for i in df_workers.index:
  for j in df_buildings.index:
    production.loc[i, j] = df_workers.loc[i, df_buildings.loc[j]].sum()

print(production.head())
#    1   2   3   4   6   5   7   8   9   10  11  12  13  14  15  16  17
# 3   9  23  36  40  16  16   4  16   9  13  13  60  11  21  16  60  13
# 4  62  28  52  29  48  48  54  62  62  36  30  50  58  50  62  50  30
# 5  14  28  30  22  42  42  10  37  14  40  22  22  40  28  37  22  22
# 6  19  21  28  24  27  27  24  16  19  28  21  30  22  15  16  30  21
# 7  25  41  12  22  57  57  14   6  25  38  17  12  44  47   6  12  17

    每栋建筑允许多个工人

    prob = pulp.LpProblem("MineColoniesProblem", pulp.LpMaximize)

# in the solved problem, assignment[i,j] == 1 whenever i is assigned to j
assignment = pulp.LpVariable.dicts("Assignment", (workers, buildings), cat="Binary")

# our objective is to maximize the sum of production
prob += sum(assignment[i][j] * production.loc[i,j]
            for i in workers for j in buildings)

# each worker can be assigned to at most one building:
for i in workers:
  prob += sum(assignment[i][j] for j in buildings) <= 1

prob.solve()
# make sure that we got an optimal solution
assert prob.status == 1

# generically, we get an integer solution
assignment_dict = {i: j for i in workers for j in buildings
                   if assignment[i][j].varValue == 1}
print(f"Total production is {prob.objective.value()}") # 1401

# here is the the solution
# assignment_dict_saved = {3: 12, 4: 1, 5: 5, 6: 16, 7: 5, 8: 6, 9: 2, 10: 17, 11: 6, 12: 10, 13: 4, 14: 6, 15: 3, 17: 7, 18: 3, 19: 3, 20: 11, 21: 12, 22: 17, 23: 15, 24: 4, 25: 10, 26: 13, 27: 9, 28: 7, 29: 7, 30: 2}

    每栋建筑最多一名工人

    prob = pulp.LpProblem("MineColoniesProblem", pulp.LpMaximize)

# in the solved problem, assignment[i,j] == 1 whenever i is assigned to j
assignment = pulp.LpVariable.dicts("Assignment", (workers, buildings), cat="Binary")

# our objective is to maximize the sum of production
prob += sum(assignment[i][j] * production.loc[i,j]
            for i in workers for j in buildings)

# each worker can be assigned to at most one building:
for i in workers:
  prob += sum(assignment[i][j] for j in buildings) <= 1

# each building has at most one worker
for j in buildings:
  prob += sum(assignment[i][j] for i in workers) <= 1

prob.solve()
# make sure that we got an optimal solution
assert prob.status == 1

# generically, we get an integer solution
assignment_dict = {i: j for i in workers for j in buildings
                   if assignment[i][j].varValue == 1}
# assignment_dict_saved = {3: 16, 4: 9, 7: 5, 8: 2, 10: 11, 11: 6, 12: 10, 14: 14, 18: 3, 19: 8, 20: 17, 21: 12, 23: 15, 24: 4, 26: 13, 27: 1, 29: 7}
print(f"Total production is {prob.objective.value()}") # 929

    我们可以看到，当我们允许每栋建筑有多个工人时，总产量会更高。这是意料之中的，因为最大化问题的约束较少。

    我们还可以将优化生产与工人随机分配到建筑物时的生产进行比较。垂直线对应于最佳生产。看起来我们做得很好。

    【讨论】：

      猜你喜欢
      • 2021-11-30
      • 1970-01-01
      • 2013-09-14
      • 2022-08-14
      • 2017-09-17
      • 1970-01-01
      • 1970-01-01
      • 1970-01-01
      • 2014-09-22
      相关资源
      最近更新 更多
      热门标签
      Java Python linux javascript Mysql C# Docker 算法 前端 SpringBoot Redis Vue spring 设计模式 .net core .net kubernetes c++ 数据库 数据结构 大数据 js 机器学习 微服务 Android Go 程序员 面试 JVM ASP.net core 云原生 人工智能 后端 PHP git CSS golang k8s Nginx Django mybatis 深度学习 多线程 React 架构 devops 爬虫 云计算 Spring Boot LeetCode