解决n皇后难题

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棋盘上的解决方案的总数，但使用n=15尝试需要一个多小时才能得到答案。有没有人可以看一下代码，给我一些加速这个程序的提示……一个python程序员新手。

#!/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

N-Queens问题的回溯算法在最坏的情况下是阶乘算法。所以对于N=8来说，8！在最坏的情况下检查解决方案的数量，N=9将其设置为9!，等等。可以看到，可能的解决方案的数量增长非常大，非常快。如果你不相信我，就去计算器那里，从1开始乘以连续的数字，告诉我计算器耗尽内存的速度有多快。

幸运的是，并不是所有可能的解决方案都必须检查。不幸的是，正确解决方案的数量仍然遵循大致的阶乘增长模式。因此，算法的运行时间以阶乘速度增长。

由于您需要找到所有正确的解决方案，因此在加速程序方面真的无能为力。在修剪搜索树中不可能的分支方面，您已经做得很好了。我不认为还有什么会产生重大影响。这只是一个很慢的算法。

Answer 2

可以使用Donald Knuth的随机估计方法来估计解决方案的数量。

从没有放置的皇后开始，下一行的允许位置数为n。随机选择一个位置，计算下一行的允许位置数(p)，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位置。

对于此行，计算下一行的允许位置数(p)，并将总解数乘以此( *=，p)。重复此操作，直到无法求解棋盘，在这种情况下，解数等于零，或者直到棋盘解出。

重复多次并计算平均解的数量(包括任何零)。这应该会给你一个快速和相当准确的近似解的数量，随着你运行的次数越多，近似值越高。

我希望这是有意义的;)

Answer 3

我建议您查看Python源代码中的test_generators.py，以获得N-Queens问题的另一种实现。

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]