Question

我刚刚解决了python中的nqueen问题。该解决方案输出了将n个皇后放置在nXn棋盘上的解决方案总数，但尝试使用n = 15需要一个多小时才能得到答案。任何人都可以看看代码并给我提示加速这个程序......一个新手python程序员。

#!/usr/bin/env python2.7

##############################################################################
# a script to solve the n queen problem in which n queens are to be placed on
# an nxn chess board in a way that none of the n queens is in check by any other
#queen using backtracking'''
##############################################################################
import sys
import time
import array

solution_count = 0

def queen(current_row, num_row, solution_list):
    if current_row == num_row:
        global solution_count 
        solution_count = solution_count + 1
    else:
        current_row += 1
        next_moves = gen_nextpos(current_row, solution_list, num_row + 1)
        if next_moves:
            for move in next_moves:
                '''make a move on first legal move of next moves'''
                solution_list[current_row] = move
                queen(current_row, num_row, solution_list)
                '''undo move made'''
                solution_list[current_row] = 0
        else:
            return None

def gen_nextpos(a_row, solution_list, arr_size):
    '''function that takes a chess row number, a list of partially completed 
    placements and the number of rows of the chessboard. It returns a list of
    columns in the row which are not under attack from any previously placed
    queen.
    '''
    cand_moves = []
    '''check for each column of a_row which is not in check from a previously
    placed queen'''
    for column in range(1, arr_size):
        under_attack =  False
        for row in range(1, a_row):
            '''
            solution_list holds the column index for each row of which a 
            queen has been placed  and using the fact that the slope of 
            diagonals to which a previously placed queen can get to is 1 and
            that the vertical positions to which a queen can get to have same 
            column index, a position is checked for any threating queen
            '''
            if (abs(a_row - row) == abs(column - solution_list[row]) 
                    or solution_list[row] == column):
                under_attack = True
                break
        if not under_attack:
            cand_moves.append(column)
    return cand_moves

def main():
    '''
    main is the application which sets up the program for running. It takes an 
    integer input,N, from the user representing the size of the chessboard and 
    passes as input,0, N representing the chess board size and a solution list to
    hold solutions as they are created.It outputs the number of ways N queens
    can be placed on a board of size NxN.
    '''
    #board_size =  [int(x) for x in sys.stdin.readline().split()]
    board_size = [15]
    board_size = board_size[0]
    solution_list = array.array('i', [0]* (board_size + 1))
    #solution_list =  [0]* (board_size + 1)
    queen(0, board_size, solution_list)
    print(solution_count)


if __name__ == '__main__':
    start_time = time.time()
    main()
    print(time.time()

Answer 1

对于N-Queens问题的回溯算法是最坏情况下的因子算法。因此对于N = 8,8！在最坏的情况下检查解决方案的数量，N = 9使其成为9！等等。可以看出，可能的解决方案的数量变得非常大，非常快。如果您不相信我，只需转到计算器并开始乘以连续数字，从1开始。让我知道计算器内存耗尽的速度。

幸运的是，并非所有可能的解决方案都必须经过检查。不幸的是，正确解决方案的数量仍然遵循大致因子增长模式。因此，算法的运行时间以阶乘的速度增长。

由于您需要找到所有正确的解决方案，因此加快程序的速度并不多。你已经在从搜索树中修剪不可能的分支方面做得很好。我不认为还有其他任何会产生重大影响的事情。这只是一个缓慢的算法。

Answer 2

可以使用Donald Knuth的随机估计方法估算解的数量。

从没有放置皇后开始，下一行的允许位置数为n。随机选择其中一个位置并计算下一行的允许位置数（p），并将其乘以n，并将其存储为解的总数（total = n * p），然后随机选择一个允许的位置

对于此行，计算下一行的允许位置数（p），并计算多个解的总数（总计* = p）。重复此操作，直到无法解决电路板，在这种情况下，解决方案的数量等于零，或直到电路板解决。

重复多次并计算平均解数（包括任意零）。这可以为您提供快速且非常准确的解决方案数量近似值，近似值可以提高您的运行次数。

我希望这是有道理的;）

Answer 3

我建议您查看Python源代码中的test_generators.py，以了解N-Queens问题的替代实现。

Python 2.6.5 (release26-maint, Sep 12 2010, 21:32:47) 
[GCC 4.4.3] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> from test import test_generators as tg
>>> n= tg.Queens(15)
>>> s= n.solve()
>>> next(s)
[0, 2, 4, 1, 9, 11, 13, 3, 12, 8, 5, 14, 6, 10, 7]
>>> next(s)
[0, 2, 4, 6, 8, 13, 11, 1, 14, 7, 5, 3, 9, 12, 10]

Answer 4

刚看到这个问题。我想提供2个解决方案。

This one returns all distinct solutions
This one returns the first valid solution found

感谢您的反馈。干杯

Answer 5

以下是我的PYTHON实施。您可能希望使用PYPY来提高速度。

通过使用O（1）时间方法来检查下一个女王是否被已经放置在棋盘上的女王攻击，有助于提高速度。

假设程序是＆＃34; nqueen.py＆＃34;，运行它的一个例子是＆＃34; python nqueen.py 6＆＃34;，其中6是6x6板的大小。

#!/bin/python
#
# TH @stackoverflow, 2016-01-20, "N Queens" with an O(1) time method to check whether the next queen is attacked
#
import sys


board_size = int(sys.argv[1])
Attacked_H  = { i:0 for i in range(0, board_size) }
Attacked_DU = { i:0 for i in range(0, board_size*2) }
Attacked_DD = { i:0 for i in range(-board_size, board_size) }


def nqueen(B, N, row, col):
    if(row >= N):
        return 1
    if(col >= N):
        print "board:\n" + str.join("\n", ["_|"*q + "Q|" + "_|"*(board_size - q - 1) for q in B])
        return 0
    B[col] = row
    if(0==(Attacked_H[row] + Attacked_DU[row+col] + Attacked_DD[row-col])):
        [Attacked_H[row], Attacked_DU[row+col], Attacked_DD[row-col]] = [ 1,1,1 ]
        nqueen(B, N, 0, col + 1)
        [Attacked_H[row], Attacked_DU[row+col], Attacked_DD[row-col]] = [ 0,0,0 ]
    nqueen(B, N, row + 1, col)


nqueen(list(range(0, board_size)), board_size, 0, 0)

Answer 6

我们也可以用遗传算法解决n-queens问题。这里描述https://youtu.be/l6qll5OldHQ，在edX课程ColumbiaX：CSMM.101x人工智能（AI）的一个讲座中。

目标函数尝试优化非攻击对的数量。下面的动画显示了R中的示例解，n = 20。有关如何使用遗传算法求解n-queens的更多详细信息，请参见：https://sandipanweb.wordpress.com/2017/03/09/solving-the-n-queen-puzzle-with-genetic-algorithm-in-r/?frame-nonce=76cf9b156a。

解决n-queen拼图

6 个答案: