八皇后问题:

在8×8格的国际象棋上摆放8个皇后，使其不能互相攻击，即任意两个皇后都不能处于同一行、同一列或同一斜线上，问有多少种摆法。正确的解有很多个,遗传算法并不直接计算一共有多少个解,而是寻找满足条件的解,从一种状态进化到一种满足8皇后不能互相攻击的状态.

Q . . . . . . .

. . . . Q . . .

. . . . . . Q .

. Q . . . . . .

. . . Q . . . .

. . . . . . . Q

. . . . . . . .

. . Q . . . . .

用matplotlib简单可视化棋盘

import random

import numpy as np

import matplotlib.pyplot as plt

def display_chessboard(coordinates):

plt.figure(figsize=(7, 7))

for i in range(9):

plt.axhline(y=i,c='black')

for i in range(9):

plt.axvline(x=i,c='black')

for x,y in coordinates:

plt.text(x+0.35,y+0.35,'Q',size=20,color='blue')

plt.xlim(0,8)

plt.ylim(0,8)

这是一个满足条件的状态

# 这是一个正确的摆法

coordinates = [(2, 0),(4, 1),(7, 2),(3, 3),(0, 4),(6, 5),(1, 6),(5, 7)]

display_chessboard(coordinates)

遗传算法求解8皇后问题

随机初始化和变异

加入部分约束条件

下面是遗传算法实现过程

1. 随机状态初始化

# 随机生成初始状态

def generate_initinal_state(length,constraint=False):

geneSet = list(range(length))

# 不同行不同列约束

if constraint:

coordinate_xs = random.sample(geneSet,length)

coordinate_ys = random.sample(geneSet,length)

coordinates = [(x,y) for x,y in zip(coordinate_xs, coordinate_ys)]

else:

coordinate_xs = [random.sample(geneSet,1)[0] for i in range(length)]

coordinate_ys = [random.sample(geneSet,1)[0] for i in range(length)]

coordinates = [(x,y) for x,y in zip(coordinate_xs, coordinate_ys)]

return coordinates

coordinates_1 = generate_initinal_state(8)

coordinates_2 = generate_initinal_state(8, True)

##### 随机初始,只有一个queen满足条件

display_chessboard(coordinates_1)

这个是没有约束的随机初始化

##### 增加行列约束后,有两个queen满足条件

display_chessboard(coordinates_2)

增加了行列约束的

2.适应度的计算

需要一个合适的适应度函数来量化当前棋盘的状态,便于跟其他状态进行比较.

逐个计算每个Queen的状态,满足条件+1,bestFitness = 8

通过Queen的坐标计算当前棋盘状态,bestFitness = 0

方案1:比较简单,就是遍历Queens,检测前后左右,对角线有无其他Queen,详细说下方案2:

Queen横坐标集合,x_set

Queen纵坐标集合.y_set

以棋盘左下角为原点,计算Queens坐标和(x+y)的集合,a_set

以棋盘右下角为原点,计算Queens坐标和(x+y)的集合,b_set

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fitness = 8 \times 4 - (len(x_{set})+len(y_{set})+len(a_{set})+len(b_{set}))fitness=8×4−(len(xset​)+len(yset​)+len(aset​)+len(bset​))

步骤1: x_set的长度就是不同列Queens的数量

步骤2: y_set的长度就是不同行Queens的数量

步骤3,4: 就是判断以Queen为中心,左右对角线有无Queen,如下图:

步骤5: 32是指8个不同x坐标,8个不同的y坐标,8个不同左右对角线,8

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8 \times 4 = 328×4=32,所以要满足不能相互攻击的条件:

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len(x_{set})+len(y_{set})+len(a_{set})+len(b_{set}) = 32len(xset​)+len(yset​)+len(aset​)+len(bset​)=32

所以在8皇后问题中,fitness = 0时最好的.

发现没有,如果Queen在同一对角线的话,坐标和是相等的,如果8个Queen的坐标和是不同的8个数字,那就满足了条件的25%了

2.1适应度函数

def get_fitness(coordinates):

x_s = [coordinate[0] for coordinate in coordinates]

y_s = [coordinate[1] for coordinate in coordinates]

x_set = set(x_s)

y_set = set(y_s)

# 左对角线

a_set = set([sum(coordinate) for coordinate in coordinates])

# 右对角线(反转x轴)

b_set = set([8 - coordinate[0]+coordinate[1] for coordinate in coordinates])

fitness = 32 -len(x_set)-len(y_set)-len(a_set)-len(b_set)

return fitness

计算一个状态的适应度

random_state = generate_initinal_state(length=8)

display_chessboard(random_state)

print('random_state fitness : ',get_fitness(random_state))

random_state fitness : 10

3.变异方法

无约束随机

这个地方可以加入一些约束,加速,但是复杂的约束会影响效率

def mutate(coordinates, coordinateSet):

index = np.random.randint(8)

new_coordinates = coordinates.copy()

picked_gene = coordinates[index]

new_gene, alternative = random.sample(coordinateSet, 2)

if picked_gene == new_gene:

new_coordinates[index] = alternative

else:

new_coordinates[index] = new_gene

return new_coordinates

从随机初始状态进化的过程

init_state = generate_initinal_state(length=8,constraint=True)

display_chessboard(init_state)

初始状态是加入了约束的,开局的状态比较理想.

print('init state fitness : ',get_fitness(init_state))

init state fitness : 4

coordinateSet = [(x,y) for x in range(8) for y in range(8)]

state_logs = []

fines_logs = []

bestState = init_state

bestFitness = get_fitness(init_state)

state_logs.append(bestState)

fines_logs.append(bestFitness)

while True:

childState = mutate(bestState,coordinateSet)

childFitness = get_fitness(childState)

state_logs.append(childState)

fines_logs.append(childFitness)

if childFitness > bestFitness:

continue

print('childState Fitness :',childFitness)

if childFitness == 0:

display_chessboard(childState)

break

bestState = childState

bestFitness = childFitness

childState Fitness : 4

childState Fitness : 4

childState Fitness : 4

childState Fitness : 4

childState Fitness : 4

childState Fitness : 3

childState Fitness : 3

childState Fitness : 3

childState Fitness : 3

childState Fitness : 3

childState Fitness : 2

childState Fitness : 2

childState Fitness : 2

childState Fitness : 2

childState Fitness : 2

childState Fitness : 2

childState Fitness : 2

childState Fitness : 2

childState Fitness : 2

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 1

childState Fitness : 0

适应度的变化

进化的过程我做成了短视频,会放到公众号: 唐牛才是食神

常驻作者: 今晚打老虎, 懂王就是懂, 唐牛才是食神

安利一下公众号:

原文链接:https://blog.csdn.net/u014281392/article/details/108670778