使用Python解决N皇后问题（编码方案）：

Question

我尝试了下面的代码。如果有任何修改可以做出来，可以降低代码的复杂性。我使用了名为board的嵌套字典。

def initialize(board,n): 
    for key in ['queen','row','col','nwtose','swtone']: 
    board[key] = {} 
    for i in range(n): 
    board['queen'][i] = -1 
    board['row'][i] = 0 
    board['col'][i] = 0 
    for i in range(-(n-1),n): 
    board['nwtose'][i] = 0 
    for i in range(2*n-1): 
    board['swtone'][i] = 0 

def printboard(board): 
    for row in sorted(board['queen'].keys()): 
    print((row,board['queen'][row])) 

def free(i,j,board): 
    return(board['queen'][i] == 0 and board['row'][i] == 0 and board['col'][j] == 0 and board['nwtose'][j-i] == 0 and board['swtone'][j+i] == 0) 

def addqueen(i,j,board): 
    board['queen'][i] = j 
    board['row'][i] = 1 
    board['col'][j] = 1 
    board['nwtose'][j-i] = 1 
    board['swtone'][j+i] = 1 

def undoqueen(i,j,board): 
    board['queen'][i] = -1 
    board['row'][i] = 0 
    board['col'][j] = 0 
    board['nwtose'][j-i] = 0 
    board['swtone'][j+i] = 0 


def placequeen(i,board): 
    n = len(board['queen'].keys()) 
    for j in range(n): 
     if free(i,j,board): 
     addqueen(i,j,board) 
     if i == n-1: 
      return(True) 
     else : 
      extendsoln = placequeen(i+1,board) 
     if extendsoln: 
      return(True) 
     else: 
      undoqueen(i,j,board) 
    else: 
     return(False) 


board = {} 
n = int(input("How many Queens? ")) 
initialize(board,n) 
if placequeen(0,board): 
    printboard(board) 

所以，当我想这个代码，我能够给输入说4，但没有得到输出。 凡excatly我要去错了???

感谢

Answer 1

当您在initialize()初始化板你board['queen'][i]设置所有值-1。你的主程序调用if placequeen(0,board):这对于j每个值调用if free(i,j,board):，检查board['queen'][i] == 0。最后一次检查失败，所有j，所以free()返回False，并且placequeen永远不会添加一个女王，因此返回False，并且您的主程序甚至不会尝试打印该板。

你的程序比需要的更复杂：我根本看不到你的字典，你可以将你的各种数组作为单独的变量。但是，您实际上并未要求简化代码。我不知道如何删除这个非印刷错误，但考虑设置board['queen'][i]到0而非-1。

我同意@RandomDavis，你需要学习如何使用调试器。我用一个来找出你所说错误的原因，但其他人可能仍然存在。