Python实现经典算法八皇后问题

目录

• 递归回溯解八皇后问题
• 遗传算法解八皇后问题

在8×8格的国际象棋上摆放八个皇后，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，问一共有多少种摆法。

分别用递归回溯算法与遗传算法实现，代码如下

递归回溯解八皇后问题

import numpy
 
def findQueen(column):
   
    if column > 7:
        global count
        count+=1
        print(matrix)        
        return
    else:
        for row in range(8):
            if check(row,column):
                matrix[row][column]=1
                arr[column]=row
                findQueen(column+1)
                matrix[row][column]=0
                arr[column]=0            
 
def check(row,column):
    for k in range(len(arr)):
        if k >= column:
            return True
        if arr[k] == row or abs(arr[k] - row) == abs(k - column):
            return False
 
if __name__ == '__main__':
    count = 0
    matrix = numpy.zeros((8,8),dtype=int)
    arr = numpy.zeros(8,dtype=int)
    findQueen(0)
    print(count)

遗传算法解八皇后问题

import copy
import random
import math
 
# 种群大小
population_size = 8
# 父种群的编码列表
parent = []
# 子种群的编码列表
children = []
# 父种群每个个体的适应度
parent_fitness = []
# 子种群每个个体的适应度
children_fitness = []
 
 
# 初始化个体
def initial_individual():
    # 个体的编码
    individual = []
    # 8个编码
    for i in range(8):
        a = random.randint(0, 7)
        individual.append(a)
    # 计算生成的个体的适应度
    fit_score = update_fitness_score(individual)
    # 加入到种群中
 
    parent_fitness.append(fit_score)
    parent.append(individual)
    return
 
 
# 更新适应度函数
def update_fitness_score(individual):
    value = 0
    for i in range(8):
        for j in range(i + 1, 8):
            if individual[i] != individual[j]:
                x = j - i
                y = abs(individual[i] - individual[j])
                if x != y:
                    value += 1
    return value
 
 
# 初始化1个种群,种群大小为population_size
def initial_population():
    for i in range(population_size):
        initial_individual()
    return
 
 
# 选择出一个父本
def select():
    # 所有个体的适应度之和
    total_score = 0
    for fit in parent_fitness:
        total_score += fit
 
    # 轮盘赌中的数
    num = random.randint(0, total_score)
    # 前面的适应度之和
    front_score = 0
    for i in range(population_size):
        front_score += parent_fitness[i]
        # 如果此时前面的适应度之和大于生成的随机数，那么该数必定落在编号为 i 的个体上
        if front_score >= num:
            return i
 
 
# 变异
def mutation(change_individual):
    # 第pos个基因发生变异
    pos = random.randint(0, 7)
    # 改变的值
    change = random.randint(0, 7)
    change_individual[pos] = change
    return change_individual
 
 
# 交叉产生后代
def hybridization():
    # 选择两个父本
    first = select()
    second = select()
    selected_parents = copy.deepcopy([parent[first], parent[second]])
    # 交换从pos1到pos2的基因
    pos1 = random.randint(0, 6)
    pos2 = random.randint(0, 6)
    # 保证pos1 <= pos2
    if pos1 > pos2:
        pos1, pos2 = pos2, pos1
    # 交叉
    tmp = selected_parents[0][pos1:pos2]
    selected_parents[0][pos1:pos2] = selected_parents[1][pos1:pos2]
    selected_parents[1][pos1:pos2] = tmp
    # 一定的概率发生变异,假设概率为0.5
    may = random.random()
    if may > 0.5:
        selected_parents[0] = mutation(selected_parents[0])
    may = random.random()
    if may > 0.5:
        selected_parents[1] = mutation(selected_parents[1])
    # 更新适应度
    first_fit = update_fitness_score(selected_parents[0])
    second_fit = update_fitness_score(selected_parents[1])
 
    # 加入到子代中
    children.append(selected_parents[0])
    children.append(selected_parents[1])
    children_fitness.append(first_fit)
    children_fitness.append(second_fit)
    return
 
 
# 初始化种群
initial_population()
# 计算迭代次数
count = 0
# not a number
find = float('nan')
while True:
    count += 1
    if count % 1000 == 0:
        print('第%d' % count + '次迭代')
    # 杂交population_size/2次产生population_size个后代
    for k in range(population_size // 2):
        hybridization()
    # 如果某个个体适应度达到28，说明此时找到了一个解
    for k in range(population_size):
        if children_fitness[k] == 28:
            # 记录解的位置
            find = k
            break
    if not math.isnan(find):
        break
    # 将子代种群放入父代中作为新的父代，子代清空
    parent[0:population_size] = children[0:population_size]
    parent_fitness[0:population_size] = children_fitness[0:population_size]
    children = []
    children_fitness = []
 
# 此时找到满足要求的子代个体
res = children[find]
print(res)
 
# 构造棋盘
res_queen = [[0 for i in range(8)] for j in range(8)]
for t in range(8):
    res_queen[res[t]][t] = 1
# 将棋盘打印
print("找到结果：")
for t in range(8):
    print(res_queen[t])